I need to find the derivative of:
$$ h(x) = \int_{0}^{x^2} (1-t^2)^{1/3} \, dt $$
Would the answer to that just be:
$$ (1-x^4)^{1/3}? $$
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6$\begingroup$ $h(x) = g(x^2)$ where $g(x) = \int_0^x (1 - t^2)^{1/3} \, dt$. What you've written down is $g'(x^2)$, which is not the same as $h'(x)$. Use the chain rule. $\endgroup$ – Qiaochu Yuan May 16 '11 at 20:56
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1$\begingroup$ See also: math.stackexchange.com/questions/6155/derivative-of-integral/… $\endgroup$ – Isaac May 16 '11 at 22:53
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Let $v = x^2$ then $$\frac{\mathrm{d}}{\mathrm{d}v} \int_0^{v} f(t) \mathrm{d}t = f(v)$$ so $$\frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x^2} f(t) \mathrm{d}t = \frac{\mathrm{d}v}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}v} \int_0^{v} f(t) \mathrm{d}t = 2x f(x^2).$$
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If $F(x)=\int_{a}^{x}f(t)\;\mathrm{d}t$, then $F^{\prime }(x)=f(x)$. By the chain rule if $F(x)=\int_{a}^{u(x)}f(t)\;\mathrm{d}t$, then
$$F^{\prime }(x)=F^{\prime }(u)u^{\prime }(x)=f(u(x))u^{\prime }(x).$$
In the present case $F(x)=h(x)$, $u(x)=x^{2}$ and $f(t)=(1-t^{2})^{1/3}$. Hence $u^{\prime }(x)=2x$ and $f(u(x))=f(x^{2})=(1-x^{4})^{1/3}$.
Thus
$$h^{\prime }(x)=2(1-x^{4})^{1/3}x.$$
A generalization is to find the derivative of an integral where both limits are functions of $x$, as in this question.
answered May 16 '11 at 21:24
