Question

How do you find the derivative of `F(x)=int ln(t+1)dt` from `[0, e^(2x)]`?

Answer 1

`F'(x) = 2e^(2x) \ ln(e^(2x)+1)`

Explanation:

If asked to find the derivative of an integral using the fundamental theorem of Calculus, you should not evaluate the integral

The Fundamental Theorem of Calculus tells us that:

`d/dx \ int_a^x \ f(t) \ dt = f(x)`

(ie the derivative of an integral gives us the original function back).

We are asked to find (notice the upper bound as changed from `x` to `x^2`)

`F'(x) = d/dx int_0^(e^(2x)) \ ln(t+1) \ dt`

Using the chain rule we can rewrite as:

`F'(x) = (d(e^(2x)))/dx d/(d(e^(2x))) int_0^(e^(2x)) \ ln(t+1) \ dt`

Now, `(d(e^(2x)))/dx = 2e^(2x)`, And, using the first result from the FTOC:

`d/(d(e^(2x))) int_0^(e^(2x)) \ ln(t+1) \ dt = ln(e^(2x)+1)`

Hence combining these trivial results we get:

`F'(x) = 2e^(2x) \ ln(e^(2x)+1)`