# How do I differentiate this integral?

That is:

$$\left(\int_{a(x)}^{b(x)}\!f(x,t)\,dt\right)'$$

I don't know how to differentiate a integral if functions of $x$ are at its limits.

Can you guys show me how to do this?

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Dear Voldemort, you might take a look at this answer. If it doesn't help, I can make additional suggestions. --- The general formula is $$\frac{d}{dt}\ g(t,\dots,t)=\frac{\partial g}{\partial x_1}(t,\dots,t)+\cdots+\frac{\partial g}{\partial x_n}(t,\dots,t).$$ – Pierre-Yves Gaillard Aug 20 '11 at 3:13
Dear "You know who", look at all the answers that were provided at math.stackexchange.com/questions/58321/… not so long ago. – Álvaro Lozano-Robledo Aug 20 '11 at 3:18
The variable $x$ occurs 3 times in your function. The recipe is: replace the first occurrence by $x_1$, the second by $x_2$, the third by $x_3$, call $g$ the new function, and apply the above formula. [I hope you will be given enough time to find the solution by yourself. --- I see that a full solution has already be given... Note there are 3 terms in the solution because of these 3 occurrences.] – Pierre-Yves Gaillard Aug 20 '11 at 3:22
Still not quite get it. In Pierre-Yves Gaillard's formula what does "g(t,...,t)" mean? Is it g(t1,...,t2)? – Voldemort Aug 20 '11 at 4:14
Put $$h(x):=\int_{a(x)}^{b(x)}\!f(x,t)\,dt.$$ You want to compute $h'(x)$. Put $$g(x_1,x_2,x_3):=\int_{a(x_1)}^{b(x_2)}\!f(x_3,t)\,dt.$$ (In particular $h(x)=g(x,x,x)$.) Then compute $$\frac{\partial g}{\partial x_1}(x_1,x_2,x_3)$$ (just differentiate with repect to $x_1$ - keeping $x_2$ and $x_3$ fixed). Then compute $$\frac{\partial g}{\partial x_1}(x,x,x)$$ (just replace $x_1$ by $x$, $x_2$ by $x$, $x_3$ by $x$). Etc. Finally: $$h'(x)=\frac{\partial g}{\partial x_1}(x,x,x)+\frac{\partial g}{\partial x_2}(x,x,x)+\frac{\partial g}{\partial x_3}(x,x,x).$$ – Pierre-Yves Gaillard Aug 20 '11 at 5:27

You can apply the following rule (Leibniz rule): [edited in response to Didier Piau's comment]:

If $$I(x)=J(u(x),v(x),x),\quad\text{with}\ J(\alpha,\beta,z)=\int_\alpha^\beta f(t,z)dt,\tag{1}$$

then, under suitable conditions, we have

$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}dt+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).\tag{2}$$

For further detais see this answer of mine.

Added 3. Pierre-Yves Gaillard's comment proves $(2)$. I derive $(2)$ as follows, by using this old blog post of mine (hereafter $t$ is the independant variable):

• (i) Let $f\left( x,t\right)$ be a real function defined in a rectangle $R=[ a,b]\times[c,d]\in\mathbb{R}^{2}$. If $f(x)$ is integrable in $x$ for each real value of $t$ and $\dfrac{\partial f\left( x,t\right) }{\partial t}$ is continuous in $x$ and $t$ in $R$, then the derivative of the following parametric integral $$I(t)=\displaystyle\int_{a}^{b}f\left( x,t\right) dx\tag{3}$$ is given by $$I^{\prime }(t)=\displaystyle\int_{a}^{b}\dfrac{\partial f\left( x,t\right) }{\partial t}dx.\tag{4}$$
• (ii) If the limits of integration are functions of $t$ $$I(t)=J(u(t),v(t),t)=\displaystyle\int_{u(t)}^{v(t)}f\left( x,t\right) dx,\tag{5}$$ then $$\dfrac{d}{dx}\displaystyle\int_{a}^{x}g\left( t\right) dt=g\left( x\right)\tag{6}$$ and $$\dfrac{d}{dx}\displaystyle\int_{x}^{b}g\left( t\right) dt=-\dfrac{d}{dx}\displaystyle\int_{b}^{x}g\left( t\right) dt=-g\left( x\right).\tag{7}$$ By the chain rule, we have $$I^{\prime }(t)=\dfrac{dI}{dt}=\dfrac{\partial J}{\partial t}\dfrac{dt}{dt}+\dfrac{\partial J}{\partial v}\dfrac{dv}{dt}+\dfrac{\partial J}{\partial u}\dfrac{du}{dt}\tag{8}$$ or

$$I^{\prime }(t)=\left( \dfrac{\partial }{\partial t}\displaystyle\int_{u}^{v}f\left( x,t\right) dx\right) \dfrac{dt}{dt}+\left( \dfrac{\partial }{\partial v}\displaystyle\int_{u}^{v}f\left( x,t\right) dx\right) \dfrac{dv\left( t\right) }{dt}+\left( \dfrac{\partial }{\partial u}\displaystyle\int_{u}^{v}f\left( x,t\right) dx\right) \dfrac{du\left( t\right) }{dt}.$$ $$\tag{9}$$ Hence $$I^{\prime }(t)=\displaystyle\int_{u\left( t\right) }^{v\left( t\right) }\dfrac{\partial f\left( x,t\right) }{\partial t}dx+f\left( v\left( t\right) ,t\right) v^{\prime }\left( t\right) -f\left( u\left( t\right) ,t\right) u^{\prime}\left( t\right)\tag{10},$$

which proves $(2)$ (with $x,t$ interchanged).

Added. Remark: for my convenience I changed the name of your functions $a(x),b(x)$. The functions $a(x),b(x)$ are the above functions $u(x),v(x)$.

Added 2. Example (here $t$ is the independant variable): If $$I(t)=\displaystyle\int_{2t}^{t^{2}}e^{tx}dx,$$

then $f\left( x,t\right) =e^{tx},u\left( t\right) =2t$ and $v\left( t\right) =t^{2}$. Hence $v^{\prime }\left( t\right) =2t, u^{\prime }\left( t\right) =2$, and $$\dfrac{\partial f\left( x,t\right) }{\partial t}=\dfrac{\partial }{\partial t}e^{tx}=xe^{tx}.$$

Thus, we get

$$I^{\prime }(t)=\displaystyle\int_{u\left( t\right) }^{v\left( t\right) }\dfrac{\partial f\left( x,t\right) }{\partial t}dx+f\left( v\left( t\right) ,t\right) v^{\prime }\left( t\right) -f\left( u\left( t\right) ,t\right) u^{\prime }\left( t\right),$$

$$I^{\prime}(t)=\displaystyle\int_{2t}^{t^{2}}xe^{tx}dx+2te^{t^{3}}-2e^{2t^{2}}=\dfrac{e^{t^{3}}\left( 3t^{3}-1\right) -e^{2t^{2}}\left( 4t^{2}-1\right)}{t^{2}}.$$

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The first displayed formula/definition is misleading precisely on the cause of the OP's trouble... The expression J(u,v,x) should be defined on its own or, a minima, replaced by J(u(x),v(x),x). This applies to the linked answer as well. – Did Aug 20 '11 at 10:50
Yes, well I would find this even more convincing if you used $J$ in your solution (possibly to write things like $\partial J/\partial\alpha$, $\partial J/\partial\beta$ and $\partial J/\partial z$). But I guess this amounts to a complete reformulation of your text to satisfy my whims... So you can just forget the whole thing. :-) – Did Aug 20 '11 at 12:58
@Américo: careful, German orthography is not so straightforward: there are Steinitz and Lipschitz, for – t.b. Aug 20 '11 at 13:14
Yes. But VfB Fortuna Chemnitz takes a t... Gottverdammt! – Did Aug 20 '11 at 13:16
@Américo: Come, come... If you or I wanted to never look stupid (sic) and blind (sic), unless we are one of those very few almost supernatural people, we should stop doing mathematics. Now. So... – Did Aug 20 '11 at 15:25

$$\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt=f(x,b(x))\frac{d}{dx}b(x)-f(x,a(x))\frac{d}{dx}a(x)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial{x}}f(x,t)dt,$$

where I have used Leibniz's Rule.

Note that if $a(x)$ and $b(x)$ are constants, then we have a special case of Leibniz's Rule.

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As @Theo said elsewhere, this is Leibniz and not Leibnitz, see en.wikipedia.org/wiki/Leibniz. – Did Aug 20 '11 at 13:01

There is an intermediate function of three variables involved, namely $$F(u,v,w):=\int_u^v f(w,t)\ dt\ .$$ One has $$F_u(u,v,w)=-f(w,u)\ ,\quad F_v(u,v,w)=f(w,v)\ ,\quad F_w(u,v,w)=\int_u^v f_w(w,t)\ dt\ ,$$ where the last formula is Leibniz' Rule "without extras".

When the variables $u$, $v$, $w$ become functions of $x$: $$u(x):=a(x)\ ,\quad v(x):=b(x)\ ,\quad w(x):=x\ ,$$ then the composition with $F$ defines a function $\phi(x):=F\bigl(a(x),b(x),x\bigr)$. In order to compute the derivative $\phi'$ we have to use the chain rule and obtain the formulas given in Nana's answer.

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