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

Mar 1, 2017

$F ' \left(x\right) = 2 {e}^{2 x} \setminus \ln \left({e}^{2 x} + 1\right)$

#### 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:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(t\right) \setminus \mathrm{dt} = f \left(x\right)$

(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 ${x}^{2}$)

$F ' \left(x\right) = \frac{d}{\mathrm{dx}} {\int}_{0}^{{e}^{2 x}} \setminus \ln \left(t + 1\right) \setminus \mathrm{dt}$

Using the chain rule we can rewrite as:

$F ' \left(x\right) = \frac{d \left({e}^{2 x}\right)}{\mathrm{dx}} \frac{d}{d \left({e}^{2 x}\right)} {\int}_{0}^{{e}^{2 x}} \setminus \ln \left(t + 1\right) \setminus \mathrm{dt}$

Now, $\frac{d \left({e}^{2 x}\right)}{\mathrm{dx}} = 2 {e}^{2 x}$, And, using the first result from the FTOC:

$\frac{d}{d \left({e}^{2 x}\right)} {\int}_{0}^{{e}^{2 x}} \setminus \ln \left(t + 1\right) \setminus \mathrm{dt} = \ln \left({e}^{2 x} + 1\right)$

Hence combining these trivial results we get:

$F ' \left(x\right) = 2 {e}^{2 x} \setminus \ln \left({e}^{2 x} + 1\right)$