Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I have a question given $f(x) = \int_1^{x^3} \sqrt{16 + t^6}dt$. The question asks to find $f^{-1'}(0)$. So I know $f^{-1'}(f(x)) = \frac{1}{f'(x)}$, so I have to solve $f(x) = 0$ or $\int_1^{x^3} \sqrt{16 + t^6}dt = 0$ first. I'm pretty sure there is a way to find the answer without having to carry out the integration because I don't have the necessary tools yet to integrate something like that, but I'm not sure how to do it.

share|cite|improve this question
up vote 1 down vote accepted

Let $G(t)$ be an antiderivative of $\sqrt{16+t^6}$ that you don't want to try to find. Neither do I, there is almost certainly no antiderivative that is expressible in terms of elementary functions. Then $$f(x)=G(x^3)-G(1).$$ Differentiate, using the Chain Rule. We get $$f'(x)=3x^2G'(x^3)=3x^2\sqrt{16+x^{18}}.$$ Continue, using information about how the derivative of an inverse function is connected to the derivative of the function.

Remark: We have used the Fundamental Theorem of Calculus, without explicitly mentioning it.

share|cite|improve this answer
+1 Thanks, I think that should be a 16 instead of a 1 underneath the square root. – hesson Dec 16 '12 at 0:47
@hesson: Thank you, I am typo-prone. Do look over things, but tend not to notice things that don't matter, like constants. – André Nicolas Dec 16 '12 at 0:49
Sorry, but I am still not sure how to use the derivative to solve $f(x) = 0$ which is needed to solve the question. – hesson Dec 16 '12 at 0:57
But the integral is from $1$ to $x^3$, so wouldn't the solution just be $x = 1$ for $f(x) = 0$? – hesson Dec 16 '12 at 1:13
Yes, I didn't look back at the equation, assumed we were starting at $0$. Indeed $f(1)=0$. – André Nicolas Dec 16 '12 at 1:16

Your Answer


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.