How do you differentiate #(e^ (2x) - e^(-2x) ) ^ 2#?

Answer 1

#4(e^{4x}-e^{-4x})#

First, differentiate "something" squared as #2xx# something. Then multiply by whatever you get by differentiating the something: #2(e^{2x}-e^{-2z})^1xx d/{dx}(e^{2x}-e^{-2x}) =2(e^{2x}-e^{-2z})xx(2e^{2x}+2e^{-2x})# (because #d/dx(e^{ax})=ae^{ax}#) #=4(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})#. Remember #(a-b)(a+b)=(a^2-b^2)# and #e^{2x}xxe^{2x}=e^{4x}#
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Answer 2

To differentiate the expression (e^(2x) - e^(-2x))^2, you can apply the chain rule. The derivative of the outer function, which is the square function, is 2 times the expression raised to the power of 1 (2*(e^(2x) - e^(-2x))^1). Then, you differentiate the inner function (e^(2x) - e^(-2x)) using the chain rule again.

The derivative of e^(2x) is 2e^(2x), and the derivative of e^(-2x) is -2e^(-2x).

So, the derivative of (e^(2x) - e^(-2x))^2 is:

2 * (e^(2x) - e^(-2x)) * (2e^(2x) + 2e^(-2x))

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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.

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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