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I am having some difficulty with this problem:

Let $I := [a,b]$ and let $f: I\to\mathbb ℝ$ be continuous on $I$. Also let $J := [c,d]$ and let $u: J\to\mathbb ℝ$ be differentiable on $J$ and satisfy $u(J)\subseteq I$. Show that if $G: J\to\mathbb ℝ$ is defined by $$ G(x) :=\int_a^{u(x)} f(t)\,dt $$ for $x$ in $J$, then $G'(x) = (f \circ u)(x)u'(x)$ for all $x\in J$.

Should I use the Leibniz rule to approach this problem?

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1 Answer 1

up vote 1 down vote accepted

Hint: $f$ is continuous, so $F(x) = \int_a^x f(t)\,dt$ is differentiable with $F' = f$. Now apply the chain rule to compute $G' = (F \circ u)'$.

Edit (in response to comment): The first claim follows from the fundamental theorem of calculus. That $G(x) = F(u(x))$ is true by definition - just plug $u(x)$ into $F$. That $G$ is differentiable follows from the fact that's it's the composition of two differentiable functions and the derivative is computed with the chain rule: $G'(x) = F'(u(x))u'(x) = f(u(x))u'(x)$.

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I am not sure how to show this explicitly. I guess I just keep thinking I am not showing enough work. Could you help show me what you mean? –  Jackson Hart Nov 28 '12 at 23:37
    
Can we use the FTC since our limits are constants? –  Jackson Hart Nov 29 '12 at 2:51

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