# Derivatives of integrals and inverse functions

## Homework Statement

Find the derivative of:

$1. f(x)=arccos(5x^3)$

$2. f(x)=∫cos(5x)sin(5t)dt$ when the integral is from 0 to x

## Homework Equations

Chain rule, dy/dx=dy/du*du/dx

## The Attempt at a Solution

For the first one, I can just take 5x^3 as u and then apply the chain rule to get

$-15x^2/\sqrt{1-5x^3}$

but maple TA keeps telling me that it's incorrect...

As for the second one, I'm not sure how to approach something like that when x is actually in the integral, as in the cos(5x). I mean, when I'm applying the chain rule...do I have to include cos in my choice of u?

Related Calculus and Beyond Homework Help News on Phys.org

## Homework Statement

Find the derivative of:

$1. f(x)=arccos(5x^3)$

$2. f(x)=∫cos(5x)sin(5t)dt$ when the integral is from 0 to x

## Homework Equations

Chain rule, dy/dx=dy/du*du/dx

## The Attempt at a Solution

For the first one, I can just take 5x^3 as u and then apply the chain rule to get

$-15x^2/\sqrt{1-5x^3}$

but maple TA keeps telling me that it's incorrect...

As for the second one, I'm not sure how to approach something like that when x is actually in the integral, as in the cos(5x). I mean, when I'm applying the chain rule...do I have to include cos in my choice of u?
Check the derivative of arccos again.
$$\frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}$$

As for the second one, when you have something like this:
$$\int_{g(x)}^{h(x)} f(t)dt$$
It's derivative is equal to
$$f(g(x)) \cdot g'(x)-f(h(x)) \cdot h'(x)$$

Ah, so the derivative of the first one should be

$-15x^2/\sqrt{1-25x^6}$?

Ah, so the derivative of the first one should be

$-15x^2/\sqrt{1-25x^6}$?
Yep!

Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant? i.e. I only have to multiply by the derivative of x?

I know what to do if I have something like

$d/dx∫cos(5t)sin(5t)dt$, where the integral is from 0 to x. I just differentiate by basically putting the x into the ts to get

$cos(5x)sin(5x)$

But when the cos5t in the integral becomes cos5x, do I just do this?

$cos(5x)sin(5x)*(d/dx)x=cos(5x)sin(5x) Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant? Yep, just take out the cos(5x) from the integral. You are left with $$f(x)=cos(5x) \int_{0}^{x} sin(5t)dt$$ You can either differentiate by the chain rule or better, solve the integral, put the limits and differentiate. Great, I finally got it, thanks again :) SammyS Staff Emeritus Science Advisor Homework Helper Gold Member Thanks. But for the second one, does it mean that I can just treat the cos(5x) as a constant? i.e. I only have to multiply by the derivative of x? I know what to do if I have something like [itex]d/dx∫cos(5t)sin(5t)dt$, where the integral is from 0 to x. I just differentiate by basically putting the x into the ts to get

$cos(5x)sin(5x)$

But when the cos5t in the integral becomes cos5x, do I just do this?

$cos(5x)sin(5x)*(d/dx)x=cos(5x)sin(5x)$
Don't forget, $\displaystyle \int_{0}^{x} \sin(5t)dt$ is a function of x, so you have to take that into account when you differentiate $\displaystyle \cos(5x)\int_{0}^{x} \sin(5t)dt\ .$