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How do you find the first and second derivative of #y = 2ln(x)#?

Answer 1

#y' = 2/x#
#y'' = -2/x^2#

We will use the product rule and the well-known derivative #(lnx)' = 1/x#.

The product rule states that for a function #f(x) = g(x) xx h(x)#, the derivative, #f'(x)#, is given by #f'(x) = g'(x) xx h(x) + h'(x) xx g(x)#.

#y' = 0 xx lnx + 2 xx 1/x#

#y' = 2/x#

We will find the second derivative using the quotient rule. The quotient rule states that for a function #f(x) = g(x)/h(x)#, the derivative is given by #f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#.

#y'' = (0 xx x - 2 xx 1)/(x)^2#

#y'' = -2/x^2#

Hopefully this helps!

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Answer 2

Adam Adkins

To find the first derivative of y = 2ln(x), you use the chain rule: dy/dx = 2(1/x) = 2/x. To find the second derivative, you differentiate the first derivative with respect to x: d^2y/dx^2 = -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.