# What are the first and second derivatives of # g(x) = cosx^2 + e^(lnx^2)ln(x)#?

This is a fairly standard chain and product rule problem.

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The first derivative of ( g(x) = \cos(x^2) + e^{\ln(x^2)} \ln(x) ) is:

[ g'(x) = -2x\sin(x^2) + \frac{2x}{x} + \frac{2e^{\ln(x^2)}}{x} + e^{\ln(x^2)} \ln(x) + \frac{e^{\ln(x^2)}}{x} ]

And the second derivative is:

[ g''(x) = -2\sin(x^2) - 4x^2\cos(x^2) + \frac{2}{x^2} + \frac{2e^{\ln(x^2)}}{x^2} + \frac{2}{x^2} + \frac{2e^{\ln(x^2)}}{x^2} + \frac{2e^{\ln(x^2)}}{x^2} + \frac{2e^{\ln(x^2)}}{x^2} + \frac{e^{\ln(x^2)}}{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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