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

Answer 1

#g'(x) =-2xsin(x^2) + 2xln(x) + x#

This is a fairly standard chain and product rule problem.

The chain rule states that: #d/dx f(g(x)) = f'(g(x))*g'(x)#

The product rule states that: #d/dx f(x)*g(x) = f'(x)*g(x) + f(g)*g'(x)#

Combining these two, we can figure out #g'(x)# easily. But first let's note that: #g(x) = cosx^2 + e^(lnx^2)ln(x) = cosx^2 + x^2ln(x)#

(Because #e^ln(x) = x# ). Now moving on to determining the derivative: #g'(x) = -2xsin(x^2) + 2xln(x) + (x^2)/x # #= -2xsin(x^2) + 2xln(x) + x#

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Answer 2

Isaiah Allen

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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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.