Question

How do you find the derivative of `g(x) = int 9*sqrt(1+t^8)dt` from 7 to `x^2`?

Answer 1

I assume you want to find `g'(x)=d/dx(\int_{7}^{x^{2}}9\sqrt{1+t^{8}}\ dt)`, where `g(x)=\int_{7}^{x^{2}}9\sqrt{1+t^{8}}\ dt`.

Let `F(x)=\int_{7}^{x}9\sqrt{1+t^{8}}\ dt` and `h(x)=x^{2}`. Note that `g(x)=F(h(x))`. By the Chain Rule, `g'(x)=F'(h(x))\cdot h'(x)`

By the Fundamental Theorem of Calculus, `F'(x)=9\sqrt{1+x^{8}}`. Since `h'(x)=2x`, the final answer is:

`g'(x)=d/dx(\int_{7}^{x^{2}}9\sqrt{1+t^{8}}\ dt)=9\sqrt{1+x^{16}}\cdot 2x=18x\sqrt{1+x^{16}}`.