# How do you differentiate #y=e^((lnx)^2)#?

Note on rewriting

Trying to rewrite is helpful in many problems. It is less helpful in this case than one might hope:

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To differentiate ( y = e^{(\ln(x))^2} ), you can use the chain rule. The chain rule states that if ( y = f(g(x)) ), then ( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) ). In this case, let ( f(u) = e^{u^2} ) and ( g(x) = \ln(x) ).

First, find the derivative of ( f(u) = e^{u^2} ), which is ( f'(u) = 2ue^{u^2} ).

Then, find the derivative of ( g(x) = \ln(x) ), which is ( g'(x) = \frac{1}{x} ).

Now, apply the chain rule:

[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) = 2(\ln(x))e^{(\ln(x))^2} \cdot \frac{1}{x} = \frac{2\ln(x)}{x}e^{(\ln(x))^2} ]

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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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