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

Solution:

Let u = x^4

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

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

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

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

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