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

# 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:

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

Using the chain rule we can rewrite as:

Hence combining these trivial results we get:

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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} ).

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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) ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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