Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|improve this question
5  
$h(x) = g(x^2)$ 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. –  Qiaochu Yuan May 16 '11 at 20:56
1  

2 Answers 2

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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