Question

how to find the derivative of the following function using Fundamental Theorem of Calculus. ?

Answer 1

You have to be a bit careful on the use of the Fundamental Theorem when you have a definite integral. Also the chain's rule apply.

Explanation:

The Fundamental Theorem of Calculus states that the derivative is the inverse function of the integral.
Reading this we would be tempted to say that

`d/dxF(x)=(2x-1)^3`.

But this is not accurate. The reason is that the integral is evaluated on `x^3` and `x^7`, and we have to take into account this. Before solving this integral let's try with a simpler one to understand the problem

`F(x)=\int_0^(x^2)1dt`.

`F(x)=t+C` to be evaluated in `x^2` and `0`, with the result

`F(x)=x^2-0=x^2`.

If we now derive `F(x)` we obtain `2x`, that it not the argument of the integral `1`.

To clarify even better, we call the indefinite integral

`I(t)=\int(2t-1)^3dt`

here we can apply the derivative safely

`I'(t)=d/dtI(t)=(2t-1)^3`.

When we pass to F(x) we have

`F(x)=I(x^7)-I(x^3)`, so the derivative has to be calculated using the chain rule:

`d/dxF(x)=I'(x^7)*d/dxx^7-I'(x^3)d/dxx^3`

We know `I'` for the fundamental theorem, so we have

`d/dxF(x)=(2(x^7)-1)^3*d/dxx^7-(2(x^3)-1)^3*d/dxx^3`

`d/dxF(x)=(2x^7-1)^3*7x^6-(2x^3-1)^3*3x^2`.