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:

$$ (1-x^4)^{1/3}? $$


Let $v = x^2$ then $$\frac{\mathrm{d}}{\mathrm{d}v} \int_0^{v} f(t) \mathrm{d}t = f(v)$$ so $$\frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x^2} f(t) \mathrm{d}t = \frac{\mathrm{d}v}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}v} \int_0^{v} f(t) \mathrm{d}t = 2x f(x^2).$$


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 $f(u(x))=f(x^{2})=(1-x^{4})^{1/3}$.


$$h^{\prime }(x)=2(1-x^{4})^{1/3}x.$$

A generalization is to find the derivative of an integral where both limits are functions of $x$, as in this question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.