What is the derivative of # lnx/(4x^2)#?

Answer 1

William Abdalla

Quotient rule, which states for #y=f(x)/g(x)#, #y'=(f'(x)g(x)-f(x)g'(x))/g(x)^2#, will help us here.

#(dy)/(dx)=((1/x)(4x^2)-lnx(8x))/(4x^4)#

#(dy)/(dx)=(4x-8xlnx)/(4x^4)=(1-2lnx)/(x^3)#

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Answer 2

Adam Adkins

To find the derivative of ( \frac{{\ln(x)}}{{4x^2}} ), we can use the quotient rule. The derivative is given by:

[ \frac{{d}}{{dx}}\left(\frac{{\ln(x)}}{{4x^2}}\right) = \frac{{\frac{{d}}{{dx}}(\ln(x)) \cdot 4x^2 - \ln(x) \cdot \frac{{d}}{{dx}}(4x^2)}}{{(4x^2)^2}} ]

[ = \frac{{\frac{{1}}{{x}} \cdot 4x^2 - \ln(x) \cdot 8x}}{{16x^4}} ]

[ = \frac{{4 - 8\ln(x)}}{{16x^3}} ]

So, the derivative is ( \frac{{4 - 8\ln(x)}}{{16x^3}} ).

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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.