# Derivative of

Using the FTOC and attemting to find the derivative of $$g(x) \int^x_0 tan^2 2tdt$$ I made $$u=tan^2$$. I know it will wind up being F(x)-F(0) but I'm at a loss with what to do with the 2t. Am I on the right track or have I chosen the wrong u?

If you could, please guide me and don't spell out the answer.

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LeonhardEuler
Gold Member
I think the problem is that there are 2 fundamental theorems of calculus. One of them is that:
$$\int_{a}^{b}f(x)dx=F(b)-F(a)$$
where F(x) is an antiderivative of f(x). The other is that
$$\frac{d}{dx}\int_{0}^{x} f(t)dt=f(x)$$
I think they mean for you to use the latter along with the product rule for differentiation.

HallsofIvy
Homework Helper
Of course, you use the product rule also because of that g(x)
but the derivative of $$\int_0^x f(t)dt$$ is just f(x).

I've gotten another hint that (tan^2 2t) should have been my u. With that I get (tan2x*tan2x) which I made (sin2x/cos2x)(sin2x/cos2x). Now I make with d/dx g(x)=d/dx[g(u)subx - g(u)sub0] the sub0 drops off because its zero and now I have d/dx[g(u)subx]=f(x) which leaves me with (sin2x/cos2x)(sin2x/cos2x).

Sorry, got lazy with the la tex.

HallsofIvy
Homework Helper
Why in the world would you change from tan(2x) to $\frac{sin(2x)}{cos(2x)}$? That just makes it more complicated!

Representing, for the moment, $\int^x_0 tan^2 2tdt$ as "f(x)", the problem is to differentiate g(x)f(x) and its derivative is, of course, g'(x)f(x)+ g(x)f'(x) (the product rule).

By the Fundamental Theorem of Calculus (one of them, anyway),
f'(x)= tan2(2x)- it's as easy as that.

$$g'(x)\int^x_0 tan^2 2tdt+ g(x)tan^2(2x)$$