Derivative of integral, which way?












1












$begingroup$


I know ${Largeint} _{x}^{sqrt x} dt$



can be written as $sqrt{x} - x$



but what is the derivative of ${Largeint} _{x}^{sqrt x} dt$?



is it 1 or derivative of $sqrt{x} - x$?






calculus integration







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    1












    $begingroup$


    I know ${Largeint} _{x}^{sqrt x} dt$



    can be written as $sqrt{x} - x$



    but what is the derivative of ${Largeint} _{x}^{sqrt x} dt$?



    is it 1 or derivative of $sqrt{x} - x$?






    calculus integration







    share|cite|improve this question







    New contributor




    user10998700 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I know ${Largeint} _{x}^{sqrt x} dt$



      can be written as $sqrt{x} - x$



      but what is the derivative of ${Largeint} _{x}^{sqrt x} dt$?



      is it 1 or derivative of $sqrt{x} - x$?






      calculus integration







      share|cite|improve this question







      New contributor




      user10998700 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I know ${Largeint} _{x}^{sqrt x} dt$



      can be written as $sqrt{x} - x$



      but what is the derivative of ${Largeint} _{x}^{sqrt x} dt$?



      is it 1 or derivative of $sqrt{x} - x$?






      calculus integration




      calculus integration






      share|cite|improve this question







      New contributor




      user10998700 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      user10998700 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question






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      asked 1 hour ago









      user10998700user10998700

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          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Second answer: derivative of $sqrt{x}-x$ which is $-1+frac{1}{2sqrt{x}}$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            It depends on with respect to which variable you're differentiating. If you're differentiating with respect to the variable $x$, then the answer is going to be $frac{1}{2sqrt{x}}-1$:



            $$
            frac{d}{dx}left(int_{x}^{sqrt x},dtright)=
            frac{d}{dx}left(sqrt{x}-xright)=
            frac{1}{2sqrt{x}}-1.
            $$



            If you're differentiating with respect to the variable $t$, then you would get $0$ because $sqrt{x}-x$ is a constant (it's like $sqrt{4}-4$ which is just a number) and the derivative of a constant, as you probably know, is $0$:



            $$
            frac{d}{dt}left(int_{x}^{sqrt x},dtright)=
            frac{d}{dt}left(sqrt{x}-xright)=
            0.
            $$






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
              $endgroup$
              – John Omielan
              1 hour ago








            • 1




              $begingroup$
              It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
              $endgroup$
              – Mike R.
              1 hour ago








            • 1




              $begingroup$
              In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
              $endgroup$
              – John Omielan
              1 hour ago












            • $begingroup$
              Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
              $endgroup$
              – Mike R.
              1 hour ago








            • 2




              $begingroup$
              This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
              $endgroup$
              – John Omielan
              53 mins ago





















            2












            $begingroup$

            The generalization of the fundamental theorem of calculus says that:



            ($int_{g(x)}^{h(x)} f(t) dt))'= f(h(x))*h'(x)-f(g(x))*g'(x)$



            (This is because of the derivative chain rule).
            It's pretty easy to go on from here.
            Good luck.






            share|cite|improve this answer










            New contributor




            Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
              $endgroup$
              – John Omielan
              1 hour ago










            • $begingroup$
              Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
              $endgroup$
              – Jonathan Perales
              1 hour ago











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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Second answer: derivative of $sqrt{x}-x$ which is $-1+frac{1}{2sqrt{x}}$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Second answer: derivative of $sqrt{x}-x$ which is $-1+frac{1}{2sqrt{x}}$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Second answer: derivative of $sqrt{x}-x$ which is $-1+frac{1}{2sqrt{x}}$






                share|cite|improve this answer









                $endgroup$



                Second answer: derivative of $sqrt{x}-x$ which is $-1+frac{1}{2sqrt{x}}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                HAMIDINE SOUMAREHAMIDINE SOUMARE

                35618




                35618























                    2












                    $begingroup$

                    It depends on with respect to which variable you're differentiating. If you're differentiating with respect to the variable $x$, then the answer is going to be $frac{1}{2sqrt{x}}-1$:



                    $$
                    frac{d}{dx}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dx}left(sqrt{x}-xright)=
                    frac{1}{2sqrt{x}}-1.
                    $$



                    If you're differentiating with respect to the variable $t$, then you would get $0$ because $sqrt{x}-x$ is a constant (it's like $sqrt{4}-4$ which is just a number) and the derivative of a constant, as you probably know, is $0$:



                    $$
                    frac{d}{dt}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dt}left(sqrt{x}-xright)=
                    0.
                    $$






                    share|cite|improve this answer











                    $endgroup$









                    • 2




                      $begingroup$
                      What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
                      $endgroup$
                      – John Omielan
                      1 hour ago








                    • 1




                      $begingroup$
                      It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 1




                      $begingroup$
                      In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
                      $endgroup$
                      – John Omielan
                      1 hour ago












                    • $begingroup$
                      Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 2




                      $begingroup$
                      This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
                      $endgroup$
                      – John Omielan
                      53 mins ago


















                    2












                    $begingroup$

                    It depends on with respect to which variable you're differentiating. If you're differentiating with respect to the variable $x$, then the answer is going to be $frac{1}{2sqrt{x}}-1$:



                    $$
                    frac{d}{dx}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dx}left(sqrt{x}-xright)=
                    frac{1}{2sqrt{x}}-1.
                    $$



                    If you're differentiating with respect to the variable $t$, then you would get $0$ because $sqrt{x}-x$ is a constant (it's like $sqrt{4}-4$ which is just a number) and the derivative of a constant, as you probably know, is $0$:



                    $$
                    frac{d}{dt}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dt}left(sqrt{x}-xright)=
                    0.
                    $$






                    share|cite|improve this answer











                    $endgroup$









                    • 2




                      $begingroup$
                      What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
                      $endgroup$
                      – John Omielan
                      1 hour ago








                    • 1




                      $begingroup$
                      It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 1




                      $begingroup$
                      In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
                      $endgroup$
                      – John Omielan
                      1 hour ago












                    • $begingroup$
                      Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 2




                      $begingroup$
                      This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
                      $endgroup$
                      – John Omielan
                      53 mins ago
















                    2












                    2








                    2





                    $begingroup$

                    It depends on with respect to which variable you're differentiating. If you're differentiating with respect to the variable $x$, then the answer is going to be $frac{1}{2sqrt{x}}-1$:



                    $$
                    frac{d}{dx}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dx}left(sqrt{x}-xright)=
                    frac{1}{2sqrt{x}}-1.
                    $$



                    If you're differentiating with respect to the variable $t$, then you would get $0$ because $sqrt{x}-x$ is a constant (it's like $sqrt{4}-4$ which is just a number) and the derivative of a constant, as you probably know, is $0$:



                    $$
                    frac{d}{dt}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dt}left(sqrt{x}-xright)=
                    0.
                    $$






                    share|cite|improve this answer











                    $endgroup$



                    It depends on with respect to which variable you're differentiating. If you're differentiating with respect to the variable $x$, then the answer is going to be $frac{1}{2sqrt{x}}-1$:



                    $$
                    frac{d}{dx}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dx}left(sqrt{x}-xright)=
                    frac{1}{2sqrt{x}}-1.
                    $$



                    If you're differentiating with respect to the variable $t$, then you would get $0$ because $sqrt{x}-x$ is a constant (it's like $sqrt{4}-4$ which is just a number) and the derivative of a constant, as you probably know, is $0$:



                    $$
                    frac{d}{dt}left(int_{x}^{sqrt x},dtright)=
                    frac{d}{dt}left(sqrt{x}-xright)=
                    0.
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 1 hour ago









                    Mike R.Mike R.

                    1,965314




                    1,965314








                    • 2




                      $begingroup$
                      What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
                      $endgroup$
                      – John Omielan
                      1 hour ago








                    • 1




                      $begingroup$
                      It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 1




                      $begingroup$
                      In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
                      $endgroup$
                      – John Omielan
                      1 hour ago












                    • $begingroup$
                      Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 2




                      $begingroup$
                      This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
                      $endgroup$
                      – John Omielan
                      53 mins ago
















                    • 2




                      $begingroup$
                      What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
                      $endgroup$
                      – John Omielan
                      1 hour ago








                    • 1




                      $begingroup$
                      It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 1




                      $begingroup$
                      In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
                      $endgroup$
                      – John Omielan
                      1 hour ago












                    • $begingroup$
                      Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
                      $endgroup$
                      – Mike R.
                      1 hour ago








                    • 2




                      $begingroup$
                      This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
                      $endgroup$
                      – John Omielan
                      53 mins ago










                    2




                    2




                    $begingroup$
                    What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
                    $endgroup$
                    – John Omielan
                    1 hour ago






                    $begingroup$
                    What is your $t$ that you are taking a derivative of? Note the variable $t$ in the integral $int_{x}^{sqrt{x}} dt$ is just a "dummy" variable as the indication of what is to be integrated, so it doesn't have any relevance outside of the integral, and with it not showing in the final result. As such, you will always get a result of $0$ for any such derivative, not just this particular example. If I'm missing something simple & obvious, I apologize and would appreciate it if you could explain. Thanks.
                    $endgroup$
                    – John Omielan
                    1 hour ago






                    1




                    1




                    $begingroup$
                    It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
                    $endgroup$
                    – Mike R.
                    1 hour ago






                    $begingroup$
                    It doesn't matter whether there is a $t$ in the expression or not. You still can differentiate with respect to anything you want. I hope I'm not doing anything wrong, am I?
                    $endgroup$
                    – Mike R.
                    1 hour ago






                    1




                    1




                    $begingroup$
                    In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
                    $endgroup$
                    – John Omielan
                    1 hour ago






                    $begingroup$
                    In that case, why not also differentiate wrt $a$, $b$, $c$, $d$, ..., with all being as valid as using $t$? You are correct $t$ doesn't need to be in the expression, but the variable should have a particular meaning. In this case, as $t$ is not defined as being anything specific, then you need to define what you mean by $t$ to make differentiating wrt to it have any meaning.
                    $endgroup$
                    – John Omielan
                    1 hour ago














                    $begingroup$
                    Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
                    $endgroup$
                    – Mike R.
                    1 hour ago






                    $begingroup$
                    Well, in the OP's example there are two variables $x$ and $t$. In my answer, I show that you get different answers depending on with respect to which of the two variables you're differentiating. That's really all there is to it. I don't know why it rubs you the wrong way.
                    $endgroup$
                    – Mike R.
                    1 hour ago






                    2




                    2




                    $begingroup$
                    This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
                    $endgroup$
                    – John Omielan
                    53 mins ago






                    $begingroup$
                    This will be my final comment, so anything we still don't agree on we'll just have to agree to disagree. Your last sentence says "If the expression under the integral sign had a $t$ in it, then there would probably be a $t$ in the answer". This is not correct. A simple example is that $int_{x}^{sqrt{x}} tdt = frac{x - x^2}{2}$. My point is that anything with $t$ would be integrated, so there will never be any $t$ in the final answer. Also, I will repeat again, what is $t$ supposed to be outside of the integral, as you can't reasonably differentiate wrt a variable if it has no meaning!
                    $endgroup$
                    – John Omielan
                    53 mins ago













                    2












                    $begingroup$

                    The generalization of the fundamental theorem of calculus says that:



                    ($int_{g(x)}^{h(x)} f(t) dt))'= f(h(x))*h'(x)-f(g(x))*g'(x)$



                    (This is because of the derivative chain rule).
                    It's pretty easy to go on from here.
                    Good luck.






                    share|cite|improve this answer










                    New contributor




                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
                      $endgroup$
                      – Jonathan Perales
                      1 hour ago
















                    2












                    $begingroup$

                    The generalization of the fundamental theorem of calculus says that:



                    ($int_{g(x)}^{h(x)} f(t) dt))'= f(h(x))*h'(x)-f(g(x))*g'(x)$



                    (This is because of the derivative chain rule).
                    It's pretty easy to go on from here.
                    Good luck.






                    share|cite|improve this answer










                    New contributor




                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$













                    • $begingroup$
                      I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
                      $endgroup$
                      – Jonathan Perales
                      1 hour ago














                    2












                    2








                    2





                    $begingroup$

                    The generalization of the fundamental theorem of calculus says that:



                    ($int_{g(x)}^{h(x)} f(t) dt))'= f(h(x))*h'(x)-f(g(x))*g'(x)$



                    (This is because of the derivative chain rule).
                    It's pretty easy to go on from here.
                    Good luck.






                    share|cite|improve this answer










                    New contributor




                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    The generalization of the fundamental theorem of calculus says that:



                    ($int_{g(x)}^{h(x)} f(t) dt))'= f(h(x))*h'(x)-f(g(x))*g'(x)$



                    (This is because of the derivative chain rule).
                    It's pretty easy to go on from here.
                    Good luck.







                    share|cite|improve this answer










                    New contributor




                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago





















                    New contributor




                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 1 hour ago









                    Jonathan PeralesJonathan Perales

                    385




                    385




                    New contributor




                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Jonathan Perales is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.












                    • $begingroup$
                      I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
                      $endgroup$
                      – Jonathan Perales
                      1 hour ago


















                    • $begingroup$
                      I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
                      $endgroup$
                      – Jonathan Perales
                      1 hour ago
















                    $begingroup$
                    I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
                    $endgroup$
                    – John Omielan
                    1 hour ago




                    $begingroup$
                    I believe you forgot to add an indication you are taking the derivative of the integral, not just evaluating the integral.
                    $endgroup$
                    – John Omielan
                    1 hour ago












                    $begingroup$
                    Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
                    $endgroup$
                    – Jonathan Perales
                    1 hour ago




                    $begingroup$
                    Thanks. You are right. Latex is a new thing to me, writing the integral itself was a tough job!
                    $endgroup$
                    – Jonathan Perales
                    1 hour ago










                    user10998700 is a new contributor. Be nice, and check out our Code of Conduct.










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                    user10998700 is a new contributor. Be nice, and check out our Code of Conduct.













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                    user10998700 is a new contributor. Be nice, and check out our Code of Conduct.
















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