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What are the first and second derivatives of #f(x)=lnx/x^2#?

Answer 1

William Abdalla

We'll use quotient rule and product rule

Using quotient rule, which states that, for a function #y=f(x)/g(x)#, #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2#

#(dy)/(dx)=((1/cancel(x))x^cancel(2)-lnx(2x))/(x^4)=(cancel(x)(1+2lnx))/x^(cancel(4)3)#

#(dy^2)/(d^2x)=((2/x)x^3-(1+2lnx)*3x^2)/x^6#

#(dy^2)/(d^2x)=(x^2(2-3(1+2lnx)))/x^6#

#(dy^2)/(d^2x)=(x^2(2-3-6lnx))/x^6#

#(dy^2)/(d^2x)=(-1-6lnx)/x^4#

#(dy^2)/(d^2x)=-(1+6lnx)/x^4#

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Answer 2

Paisley Johnson

The first derivative of ( f(x) = \frac{\ln x}{x^2} ) is ( f'(x) = \frac{1 - 2\ln x}{x^3} ).

The second derivative of ( f(x) = \frac{\ln x}{x^2} ) is ( f''(x) = \frac{2\ln x - 6}{x^4} ).

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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.