# Differentiation with respect to integral boundary

Using the chain rule show the following proposition:

Let $f$ be continuously on $[a,b]$ and $g:J\to[a,b]$ continuously differentiable for an interval $J$. We write

$$H(x)=\int\limits_{a}^{g(x)}f(t)\,\mathrm dt,\qquad x\in J.$$

$H$ is differentiable and one has $$H'(x)=f\left(g\left(x\right)\right)g'(x).$$

After proving the correctness of the proposition use it to compute the derivative of $$H(x)=\int\limits_{1}^{\exp(x)}\ln(2t)\,\mathrm dt.$$

Proof. Let $h\in\mathbb{R}$ with $h>0$ and $x+h\in[a,b]$. Then we know that

\begin{align} \frac{H(x+h)-H(x)}{h} &= \frac{1}{h}\left(\int\limits_{a}^{g(x+h)}f(t)\,\mathrm dt - \int\limits_{a}^{g(x)}f(t)\,\mathrm dt\right)\\ &= \frac{1}{h}\int\limits_{g(x)}^{g(x+h)}f(t)\,\mathrm dt\\ &\overset{(1)}{=} \frac{1}{h}\int\limits_{x}^{x+h}f(g(t))g'(t)\,\mathrm dt = \frac{1}{h}\int\limits_{x}^{x+h}H'(t)\,\mathrm dt\\ &\overset{(2)}{=} \frac{1}{h}\cdot h\cdot H'(\xi)=H'(\xi),\qquad \xi\in[x,x+h]. \end{align}

In $(1)$ we just use backwards substitution for integration to move $g$ out of the integral boundary into the integrand. $(2)$ is more tricky because we know that for every continuously function $f:[a,b]\to\mathbb{R}$ there exists one $\xi\in[a,b]$ with $$\int\limits^b_af(x)\,\mathrm dx=(b-a)f(\xi).$$ In our case it is obvious that $(x+h)-x=h$. Now we see that $h\to 0$ yields $\xi\to x$ and therefore $$\frac{H(x+h)-H(x)}{h}\longrightarrow H'(x).$$ The same principle applies for $h<0$ and $h\to 0$.$\quad\square$

Example. Let now $$H(x)=\int\limits_{1}^{\exp(x)}\ln(2t)\,\mathrm dt$$ then $f(t)=\ln(2t),\,g(t)=\exp(t)$. This yields

\begin{align} H'(x)&=\ln(2\exp(x))\cdot\exp(x)=\exp(x)(x+\ln(2)). \end{align}

I would like to know whether my approach is correct and whether I could simplify some steps in there, as I usually do everything more complicated than neccessary.

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I don't know if I am making things too easy, but as we have $f$ continuous on $[a,b]$ we can use the fundamental theorem of calculus and obtain: $$H(x) = \int_a^{g(x)} f(t)\, dt = F(g(x))-F(a)$$ Differentiating H w.r.t x we get $$H'(x) = (F(g(x))-F(a))' = F'(g(x))\cdot g'(x) = f(g(x))\cdot g'(x)$$ as we know that $g(x)$ is continuously differentiable. – k1next Jan 7 '13 at 13:22
@macydanim: Holy shit... this looks awesome. I will write this down as a possible alternative solution. – Christian Ivicevic Jan 7 '13 at 13:28
glad I could help ;) – k1next Jan 7 '13 at 14:06

While the proof in the question is correct, it is somewhat redundant. The beginning of the proof is something already done in the proof of the Fundamental Theorem. So it makes perfect sense to use the FTC, as macydanim has done: let $F$ be an antiderivative of $f$; then $H(x)=F(g(x))-F(a)$, and finally $$H'(x) = (F(g(x))-F(a))' = F'(g(x))\cdot g'(x) = f(g(x))\cdot g'(x)$$