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

1 Answer
Mar 1, 2017

Answer:

# 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) #