# Thread: Fundemental Therom of Calculus with Chain Rule

1. ## Fundemental Therom of Calculus with Chain Rule

Find $\frac{d}{dx}\int^{x^4}_1 sec t, dt$

Solution:

Let u = x^4

$\frac{d}{dx}\int^{x^4}_1 sec t, dt = \frac{d}{dx}\int^u_1 sect, dt$

$= \frac{d}{du}(\int^{u}_1 sec t,dt) \frac{du}{dx}$

$= sec u \frac{du}{dx}$

$= sec({x^4})*{4x^3}$

Maybe I'm just rusty on the chain rule, I thought I needed to take the derivative of sec and then the derivative of what's inside the brackets: -tanxsecx * the derivative of u or t or whatever it is that would be inside the brackets.

Could someone give me a rough walk through of how the textbook came up with the solution that it did?

Thanks,

Graeme

2. Originally Posted by Glorzifen
Find $\frac{d}{dx}\int^{x^4}_1 sec t, dt$

Solution:

Let u = x^4

$\frac{d}{dx}\int^{x^4}_1 sec t, dt = \frac{d}{dx}\int^u_1 sect, dt$

$= \frac{d}{du}(\int^{u}_1 sec t,dt) \frac{du}{dx}$

$= sec u \frac{du}{dx}$

$= sec({x^4})*{4x^3}$

Maybe I'm just rusty on the chain rule, I thought I needed to take the derivative of sec and then the derivative of what's inside the brackets: -tanxsecx * the derivative of u or t or whatever it is that would be inside the brackets.

Could someone give me a rough walk through of how the textbook came up with the solution that it did?

Thanks,

Graeme
The conclusion of the Fundemental solution of calculus is

let $g(x)=\int_{a}^{x}f(t)dt$ then

$\frac{d}{dx}g(x)=\frac{d}{dx} \int_{a}^{x}f(t)dt=f(x)$

Now they are just using the chain rule.

i.e $u=x^4$

Then

$g(u)=\int_{a}^{u}f(t)dt$

$\frac{d}{dx}g(u)=g'(u)\frac{du}{dx}=4x^3f(u)$