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I am trying to show that $$ \frac{d}{dx} \int\limits^{a(x)}_{0} f(x,y)dy = \int\limits^{a}_{0} \frac{\partial f}{\partial x}dy + a'(x)f(x,a) $$

I know this has something to do with the fundamental theorem of calculus but am having trouble making any sort of progress. If someone could point me in the right direction, it would be much appreciated.

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See the more general question How do I differentiate this integral? The integral is $$\int_{a(x)}^{b(x)}\!f(x,t)\,dt.$$ –  Américo Tavares Jan 25 '13 at 16:30

4 Answers 4

up vote 1 down vote accepted

Let a $C^1$ function $f$ and $$ F(x,u,v,t)=\int_{u}^{v}f(x,t) dx. $$ Use the implicit differentiation ruler in function $$ \varphi(x)=F(x,u(x),v(x),t(x))= \int_{u(x)}^{v(x)}f(x,t(x)) dx . $$

Here, $u=u(x)$, $v=v(x)$, $t=t(x)$ are $C^1$ functions.

Implicit differentiation rule: For all fix $x_0$ we have $u_0=u(x_0)$, $v_0=v(x_0)$, $t_0=t(x_0)$ and

\begin{align} \left.\frac{d\varphi(x)}{dx}\right|_{x=x_0}= & \left.\frac{\partial F(x,u_0,v_0,t_0)}{\partial x}\right|_{x=x_0} \\ + & \left.\frac{\partial F (x_0,u(x),v_0,t_0)}{\partial x}\right|_{x=x_0} \\ + & \left.\frac{\partial F (x_0,u(x),v_0,t_0)}{\partial x}\right|_{x=x_0} \\ + & \left.\frac{\partial F (x_0,u_0,v(x),t_0)}{\partial x}\right|_{x=x_0} \\ + & \left.\frac{\partial F (x_0,u_0,v_0,t(x))}{\partial x}\right|_{x=x_0} \\ \end{align}

By chain rule, we get: \begin{align} \left.\frac{d\varphi(x)}{dx}\right|_{x=x_0}= & \left(\left.\frac{\partial F(x,u_0,v_0,t_0)}{\partial x}\right|_{x=x_0}\right) \cdot \left(\left.\frac{dx}{dx}\right|_{x=x_0}\right) \\ + & \left(\left.\frac{\partial F (x_0,u,v_0,t_0)}{\partial u}\right|_{u=u_0}\right) \cdot \left(\left.\frac{du}{dx}\right|_{x=x_0}\right) \\ + & \left(\left.\frac{\partial F (x_0,u_0,v,t_0)}{\partial v}\right|_{v=v_0}\right) \cdot \left(\left.\frac{dv}{dx}\right|_{x=x_0}\right) \\ + & \left(\left.\frac{\partial F (x_0,u_0,v_0,t)}{\partial t}\right|_{t=t_0}\right) \cdot \left(\left.\frac{dt}{dx}\right|_{x=x_0}\right) \end{align} Then the result fellowing by Leibniz rule and fundametal teorem of calculus.

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We set $$ F(x,y)=\int_0^yf(x,s)\,ds, $$ and assume that $\partial_1F$ and $\partial_2F$ exist. This is the case, e.g. when $f$ is continuous in $y$ and differentiable in $x$. Then $$ \partial_1F(x,y)=\int_0^y\frac{\partial f}{\partial x}(x,s)\,ds,\quad \partial_2F(x,y)=f(x,y). $$ (See, e.g. Prove that $h'(t)=\int_{a}^{b}\frac{\partial\phi}{\partial t}ds$. for the first identity). It follows that \begin{eqnarray} \frac{d}{dx}\int_0^{a(x)}f(x,y)\,dy&=&\frac{d}{dx}(F(x,a(x)))=\partial_1F(x,a(x))+a'(x)\partial_2F(x,a(x)\\ &=&\int_0^{a(x)}\frac{\partial f}{\partial x}(x,y)\,dy+a'(x)f(x,a(x)). \end{eqnarray}

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I'm having the same problem with this as I am with Alex's answer. Shouldn't the derivative of the integral be $\frac{d}{dx}(F(x,a(x)) - F(x,0))$? Using that and the chain rule, I get $\frac{\delta}{\delta x}F(x,a(x)) a'(x) - \frac{\delta}{\delta x}F(x,0)$. I am also not sure where you get $\delta_1 F(x,a(x))$ from? –  rioneye Jan 25 '13 at 14:29
    
@rioneye After all it should be $F (x, 0)\cdot \frac{\mathrm d 0}{\mathrm d x}=0$ right? –  uforoboa Jan 25 '13 at 15:56
    
Using only the definition we have $F(x,0)=\int_0^0f(x,s)\,ds=0$. –  Mercy Jan 25 '13 at 17:38

This is Leibniz's integration rule. The proof is by partial differentiation. Define the function $F(x,z):=\int_0^z f(x,y)dy$. Then your integral becomes $F(x,a(x))$. By partial differentiation, we get:

$$\frac{dF}{dx}=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{dz}{dx}$$

Now, $\frac{\partial F}{\partial x}=\int_0^z \frac{\partial f}{\partial x}$. As well, $\frac{\partial F}{\partial z}=f(x,y)dy$ by fundamental theorem of calculus. Finally, writing $z=a(x)$, we have $\frac{dz}{dx}=a'(x)$ giving the desired result.

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Doesn't the integral become $F(x,a(x)) - F(x,0)$? If I go by your example shouldn't the next part be $\frac{d F}{dx}=\frac{\delta F}{\delta z} \frac{dz}{dx}$ via the chain rule? I'm not sure where adding the $\frac{\delta F}{\delta x}$ came from. Also, thanks for the link to the Leibniz rule page. –  rioneye Jan 25 '13 at 4:26
    
No, the integral is just $F(x,a(x))$, as $F(x,0)=0$. Moreover, you are differentiating in the parameter $x$ on which both inputs in $F$ depend on. –  Alex R. Jan 25 '13 at 19:06

Hint: Define $H(x,z)=\int_0^{z} f(x,y)dy$, hence you want to calculate $\displaystyle\frac{d H(x,z(x))}{dx}$ where $z(x)=a(x)$

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