What's the intuition behind this chain rule usage in the fundamental theorem of calc?












3












$begingroup$


I understand this:



enter image description here



but I don't understand this example fully:



enter image description here



So I get the intuition behind the idea that the sliver at some point x in the area function (g(x)) is just the y coordinate of the original function (f(x)), but I'm sure why we need chain rule in the example. Can someone show me an easier example that might pump my intuition?



Like say the example was this instead:



$$ frac{d}{dx} int_{a}^{x^2} t $$



So the original function t is just a line with a slope of 1 going up at a 45 degree angle. (1,1) and (2,2) are points on $t$. Using chain rule, the derivative of this graph is going to be:



$$x^2 cdot 2x = 2x^3$$ right?



And I guess the intuition behind this is that the upper limit (x^2) is increasing exponentially relative to x and so that needs to be taken into account somehow. Chain rule is that way? It's just strange to me that the area isn't just increasing by $$x^2$$ but by $$x^2 cdot 2x$$






integration







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  • $begingroup$
    Let $F(x)=int_a^{g(x)} f(t),dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=int_{g(x)}^{g(x+h)}f(t),dt$. Using the mean value theorem, there is a number $xiin[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $hto0$. Can you finish? Does this provide the intuition you covet?
    $endgroup$
    – Mark Viola
    Dec 30 '18 at 20:27












  • $begingroup$
    nah I don't see it. I liked the accepted answer better.
    $endgroup$
    – Jwan622
    Jan 4 at 14:47










  • $begingroup$
    If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $hto0$ you will get$f(g(x)),times, g'(x)$
    $endgroup$
    – Mark Viola
    Jan 4 at 17:24


















3












$begingroup$


I understand this:



enter image description here



but I don't understand this example fully:



enter image description here



So I get the intuition behind the idea that the sliver at some point x in the area function (g(x)) is just the y coordinate of the original function (f(x)), but I'm sure why we need chain rule in the example. Can someone show me an easier example that might pump my intuition?



Like say the example was this instead:



$$ frac{d}{dx} int_{a}^{x^2} t $$



So the original function t is just a line with a slope of 1 going up at a 45 degree angle. (1,1) and (2,2) are points on $t$. Using chain rule, the derivative of this graph is going to be:



$$x^2 cdot 2x = 2x^3$$ right?



And I guess the intuition behind this is that the upper limit (x^2) is increasing exponentially relative to x and so that needs to be taken into account somehow. Chain rule is that way? It's just strange to me that the area isn't just increasing by $$x^2$$ but by $$x^2 cdot 2x$$






integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    Let $F(x)=int_a^{g(x)} f(t),dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=int_{g(x)}^{g(x+h)}f(t),dt$. Using the mean value theorem, there is a number $xiin[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $hto0$. Can you finish? Does this provide the intuition you covet?
    $endgroup$
    – Mark Viola
    Dec 30 '18 at 20:27












  • $begingroup$
    nah I don't see it. I liked the accepted answer better.
    $endgroup$
    – Jwan622
    Jan 4 at 14:47










  • $begingroup$
    If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $hto0$ you will get$f(g(x)),times, g'(x)$
    $endgroup$
    – Mark Viola
    Jan 4 at 17:24
















3












3








3





$begingroup$


I understand this:



enter image description here



but I don't understand this example fully:



enter image description here



So I get the intuition behind the idea that the sliver at some point x in the area function (g(x)) is just the y coordinate of the original function (f(x)), but I'm sure why we need chain rule in the example. Can someone show me an easier example that might pump my intuition?



Like say the example was this instead:



$$ frac{d}{dx} int_{a}^{x^2} t $$



So the original function t is just a line with a slope of 1 going up at a 45 degree angle. (1,1) and (2,2) are points on $t$. Using chain rule, the derivative of this graph is going to be:



$$x^2 cdot 2x = 2x^3$$ right?



And I guess the intuition behind this is that the upper limit (x^2) is increasing exponentially relative to x and so that needs to be taken into account somehow. Chain rule is that way? It's just strange to me that the area isn't just increasing by $$x^2$$ but by $$x^2 cdot 2x$$






integration







share|cite|improve this question









$endgroup$




I understand this:



enter image description here



but I don't understand this example fully:



enter image description here



So I get the intuition behind the idea that the sliver at some point x in the area function (g(x)) is just the y coordinate of the original function (f(x)), but I'm sure why we need chain rule in the example. Can someone show me an easier example that might pump my intuition?



Like say the example was this instead:



$$ frac{d}{dx} int_{a}^{x^2} t $$



So the original function t is just a line with a slope of 1 going up at a 45 degree angle. (1,1) and (2,2) are points on $t$. Using chain rule, the derivative of this graph is going to be:



$$x^2 cdot 2x = 2x^3$$ right?



And I guess the intuition behind this is that the upper limit (x^2) is increasing exponentially relative to x and so that needs to be taken into account somehow. Chain rule is that way? It's just strange to me that the area isn't just increasing by $$x^2$$ but by $$x^2 cdot 2x$$






integration




integration






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share|cite|improve this question











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asked Dec 30 '18 at 18:40









Jwan622Jwan622

2,40111632




2,40111632












  • $begingroup$
    Let $F(x)=int_a^{g(x)} f(t),dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=int_{g(x)}^{g(x+h)}f(t),dt$. Using the mean value theorem, there is a number $xiin[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $hto0$. Can you finish? Does this provide the intuition you covet?
    $endgroup$
    – Mark Viola
    Dec 30 '18 at 20:27












  • $begingroup$
    nah I don't see it. I liked the accepted answer better.
    $endgroup$
    – Jwan622
    Jan 4 at 14:47










  • $begingroup$
    If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $hto0$ you will get$f(g(x)),times, g'(x)$
    $endgroup$
    – Mark Viola
    Jan 4 at 17:24




















  • $begingroup$
    Let $F(x)=int_a^{g(x)} f(t),dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=int_{g(x)}^{g(x+h)}f(t),dt$. Using the mean value theorem, there is a number $xiin[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $hto0$. Can you finish? Does this provide the intuition you covet?
    $endgroup$
    – Mark Viola
    Dec 30 '18 at 20:27












  • $begingroup$
    nah I don't see it. I liked the accepted answer better.
    $endgroup$
    – Jwan622
    Jan 4 at 14:47










  • $begingroup$
    If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $hto0$ you will get$f(g(x)),times, g'(x)$
    $endgroup$
    – Mark Viola
    Jan 4 at 17:24


















$begingroup$
Let $F(x)=int_a^{g(x)} f(t),dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=int_{g(x)}^{g(x+h)}f(t),dt$. Using the mean value theorem, there is a number $xiin[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $hto0$. Can you finish? Does this provide the intuition you covet?
$endgroup$
– Mark Viola
Dec 30 '18 at 20:27






$begingroup$
Let $F(x)=int_a^{g(x)} f(t),dt$ where $f$ is continuous and $g$ is differentiable. Then, $F(x+h)-F(x)=int_{g(x)}^{g(x+h)}f(t),dt$. Using the mean value theorem, there is a number $xiin[g(x),g(x+h)]$ so that $$F(x+h)-F(x)=f(xi) (g(x+h)-g(x))$$ Now divide by $h$ and let $hto0$. Can you finish? Does this provide the intuition you covet?
$endgroup$
– Mark Viola
Dec 30 '18 at 20:27














$begingroup$
nah I don't see it. I liked the accepted answer better.
$endgroup$
– Jwan622
Jan 4 at 14:47




$begingroup$
nah I don't see it. I liked the accepted answer better.
$endgroup$
– Jwan622
Jan 4 at 14:47












$begingroup$
If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $hto0$ you will get$f(g(x)),times, g'(x)$
$endgroup$
– Mark Viola
Jan 4 at 17:24






$begingroup$
If you don't understand this, I suggest you have another look and try your best. It is a very basic development. When you divide by $h$ and let $hto0$ you will get$f(g(x)),times, g'(x)$
$endgroup$
– Mark Viola
Jan 4 at 17:24












2 Answers
2






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3












$begingroup$

I will use the notation in your screenshot. If you understand that $g(x) = int_a^x f(t) , dt$ has derivative $g'(x) = f(x)$, then $int_a^{h(x)} f(t) , dt = g(h(x))$ has derivative $g'(h(x)) h'(x)$ by the chain rule.



I think any more intuitive explanation about rate of change of area would be more generally explained by whatever intuitive explanation you have for why the chain rule works in general.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    In general, let $H'(t)=h(t)$, then:
    $$int_{f(x)}^{g(x)} h(t)dt=H(t)|_{f(x)}^{g(x)}=H(g(x))-H(f(x)) Rightarrow \
    begin{align}left(int_{f(x)}^{g(x)} h(t)dtright)'_x&=[H(g(x))-H(f(x))]'_x =\
    &=h(g(x))cdot (g(x))'_x-h(f(x))cdot (f(x))'_x.end{align}$$
    Since $f(x)=1,g(x)=x^4, h(t)=sec t$, then:
    $$left(int_{1}^{x^4} sec t dtright)'_x=h(x^4)cdot (x^4)'_x-h(1)cdot (1)'_x=sec x^4cdot 4x^3.$$
    Source: See Variable limits form section at Leibniz integral rule
    .






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      3












      $begingroup$

      I will use the notation in your screenshot. If you understand that $g(x) = int_a^x f(t) , dt$ has derivative $g'(x) = f(x)$, then $int_a^{h(x)} f(t) , dt = g(h(x))$ has derivative $g'(h(x)) h'(x)$ by the chain rule.



      I think any more intuitive explanation about rate of change of area would be more generally explained by whatever intuitive explanation you have for why the chain rule works in general.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        I will use the notation in your screenshot. If you understand that $g(x) = int_a^x f(t) , dt$ has derivative $g'(x) = f(x)$, then $int_a^{h(x)} f(t) , dt = g(h(x))$ has derivative $g'(h(x)) h'(x)$ by the chain rule.



        I think any more intuitive explanation about rate of change of area would be more generally explained by whatever intuitive explanation you have for why the chain rule works in general.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          I will use the notation in your screenshot. If you understand that $g(x) = int_a^x f(t) , dt$ has derivative $g'(x) = f(x)$, then $int_a^{h(x)} f(t) , dt = g(h(x))$ has derivative $g'(h(x)) h'(x)$ by the chain rule.



          I think any more intuitive explanation about rate of change of area would be more generally explained by whatever intuitive explanation you have for why the chain rule works in general.






          share|cite|improve this answer









          $endgroup$



          I will use the notation in your screenshot. If you understand that $g(x) = int_a^x f(t) , dt$ has derivative $g'(x) = f(x)$, then $int_a^{h(x)} f(t) , dt = g(h(x))$ has derivative $g'(h(x)) h'(x)$ by the chain rule.



          I think any more intuitive explanation about rate of change of area would be more generally explained by whatever intuitive explanation you have for why the chain rule works in general.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 18:47









          angryavianangryavian

          42.7k23482




          42.7k23482























              1












              $begingroup$

              In general, let $H'(t)=h(t)$, then:
              $$int_{f(x)}^{g(x)} h(t)dt=H(t)|_{f(x)}^{g(x)}=H(g(x))-H(f(x)) Rightarrow \
              begin{align}left(int_{f(x)}^{g(x)} h(t)dtright)'_x&=[H(g(x))-H(f(x))]'_x =\
              &=h(g(x))cdot (g(x))'_x-h(f(x))cdot (f(x))'_x.end{align}$$
              Since $f(x)=1,g(x)=x^4, h(t)=sec t$, then:
              $$left(int_{1}^{x^4} sec t dtright)'_x=h(x^4)cdot (x^4)'_x-h(1)cdot (1)'_x=sec x^4cdot 4x^3.$$
              Source: See Variable limits form section at Leibniz integral rule
              .






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                In general, let $H'(t)=h(t)$, then:
                $$int_{f(x)}^{g(x)} h(t)dt=H(t)|_{f(x)}^{g(x)}=H(g(x))-H(f(x)) Rightarrow \
                begin{align}left(int_{f(x)}^{g(x)} h(t)dtright)'_x&=[H(g(x))-H(f(x))]'_x =\
                &=h(g(x))cdot (g(x))'_x-h(f(x))cdot (f(x))'_x.end{align}$$
                Since $f(x)=1,g(x)=x^4, h(t)=sec t$, then:
                $$left(int_{1}^{x^4} sec t dtright)'_x=h(x^4)cdot (x^4)'_x-h(1)cdot (1)'_x=sec x^4cdot 4x^3.$$
                Source: See Variable limits form section at Leibniz integral rule
                .






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In general, let $H'(t)=h(t)$, then:
                  $$int_{f(x)}^{g(x)} h(t)dt=H(t)|_{f(x)}^{g(x)}=H(g(x))-H(f(x)) Rightarrow \
                  begin{align}left(int_{f(x)}^{g(x)} h(t)dtright)'_x&=[H(g(x))-H(f(x))]'_x =\
                  &=h(g(x))cdot (g(x))'_x-h(f(x))cdot (f(x))'_x.end{align}$$
                  Since $f(x)=1,g(x)=x^4, h(t)=sec t$, then:
                  $$left(int_{1}^{x^4} sec t dtright)'_x=h(x^4)cdot (x^4)'_x-h(1)cdot (1)'_x=sec x^4cdot 4x^3.$$
                  Source: See Variable limits form section at Leibniz integral rule
                  .






                  share|cite|improve this answer









                  $endgroup$



                  In general, let $H'(t)=h(t)$, then:
                  $$int_{f(x)}^{g(x)} h(t)dt=H(t)|_{f(x)}^{g(x)}=H(g(x))-H(f(x)) Rightarrow \
                  begin{align}left(int_{f(x)}^{g(x)} h(t)dtright)'_x&=[H(g(x))-H(f(x))]'_x =\
                  &=h(g(x))cdot (g(x))'_x-h(f(x))cdot (f(x))'_x.end{align}$$
                  Since $f(x)=1,g(x)=x^4, h(t)=sec t$, then:
                  $$left(int_{1}^{x^4} sec t dtright)'_x=h(x^4)cdot (x^4)'_x-h(1)cdot (1)'_x=sec x^4cdot 4x^3.$$
                  Source: See Variable limits form section at Leibniz integral rule
                  .







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 19:39









                  farruhotafarruhota

                  22.2k2942




                  22.2k2942






























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