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How do you find the derivative for #ln(e^(2x))#?

Answer 1

Anna A. shows the "easy way" to answer this question, but it is often helpful to see other ways of answering a question.

What if you "messed up" and didn't simplify first?

No problem, you'll still get the correct answer.

We'll use #d/(dx)(ln(g(x))) = 1/g(x) g'(x)#

And, because, in this problem, #g(x)= e^(2x)#, we'll also use: #d/(dx)(e^(2x)) = 2e^(2x)#

So, here we go:

#d/(dx)(ln(e^(2x))) = 1/e^(2x) [2e^(2x)] = (2e^(2x))/e^(2x)#,

but this answer simplifies to:

#d/(dx)(ln(e^(2x))) = (2e^(2x))/e^(2x) = 2#,

We got the same answer. We just took a different route to get there.

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Answer 2

Christopher Allers

To find the derivative of ln(e^(2x)), you can use the chain rule. The derivative of ln(u) with respect to x is (1/u) * du/dx. In this case, u = e^(2x). The derivative of e^(2x) with respect to x is 2e^(2x). Therefore, the derivative of ln(e^(2x)) is (1/e^(2x)) * 2e^(2x) = 2.

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Answer from HIX Tutor

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.