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

Question

Paisley Johnson · Accepted Answer

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

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) #

Evelyn Agard · Answer

undefined To find the derivative of $ F(x) = \int_{0}^{e^{2x}} \ln(t + 1) \, dt $, you can use the Fundamental Theorem of Calculus along with the chain rule. The derivative is given by $ F'(x) = \ln(e^{2x} + 1) \cdot 2e^{2x} $.

Paisley Johnson · Answer

undefined To find the derivative of $ F(x) = \int_{0}^{e^{2x}} \ln(t + 1) dt $, you can use the Fundamental Theorem of Calculus and the Chain Rule.

First, apply the Fundamental Theorem of Calculus, which states that if $ F(x) = \int_{a}^{g(x)} f(t) dt $, then $ F'(x) = f(g(x)) \cdot g'(x) $.

In this case, $ f(t) = \ln(t + 1) $ and $ g(x) = e^{2x} $.

Now, find the derivative of $ g(x) $ with respect to $ x $, which is $ g'(x) = 2e^{2x} $.

Then, substitute these values into the Fundamental Theorem of Calculus:

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

So, the derivative of $ F(x) $ is:

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