Struggling with the FTOC: Derivative of g(x)

[Echo 6 Sierra]

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.

 

[LeonhardEuler]

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]

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).

 

[Echo 6 Sierra]

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]

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.

So your derivative is
$$g'(x)\int^x_0 tan^2 2tdt+ g(x)tan^2(2x)$$

 

[Echo 6 Sierra]

...good grief...I have a long way to go. Thank you.