What is the derivative of $f (x) = \frac{x^{3} - {(ln x)}^{2}}{{ln x}^{2}}$ $f(x) = (x^3-(lnx)^2)/(lnx^2)$ ?

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Answer 1 · 2016-04-27T11:12:40

Use quotent rule and chain rule. Answer is:

$f'(x)=(3x^3lnx^2-2(lnx)^2-2x^3)/(x(lnx^2)^2)$

This is a simplified version. See Explanation to watch until which point it can be accepted as a derivative.

$f(x)=(x^3-(lnx)^2)/lnx^2$

$f'(x)=((x^3-(lnx)^2)'*lnx^2-(x^3-(lnx)^2)(lnx^2)')/(lnx^2)^2$

$f'(x)=((3x^2-2lnx*(lnx)')*lnx^2-(x^3-(lnx)^2)1/x^2(x^2)')/(lnx^2)^2$

$f'(x)=((3x^2-2lnx*1/x)*lnx^2-(x^3-(lnx)^2)1/x^2*2x)/(lnx^2)^2$

At this form, it is actually acceptable. But to further simplify it:

$f'(x)=((3x^2-2lnx/x)*lnx^2-(x^3-(lnx)^2)2/x)/(lnx^2)^2$

$f'(x)=(3x^2lnx^2-2lnx/xlnx^2-x^3*2/x+(lnx)^2*2/x)/(lnx^2)^2$

$f'(x)=(3x^3lnx^2-2lnxlnx^2-x^3*2+(lnx)^2*2)/(x(lnx^2)^2)$

$f'(x)=(3x^3lnx^2-4(lnx)^2-2x^3+2(lnx)^2)/(x(lnx^2)^2)$

$f'(x)=(3x^3lnx^2-2(lnx)^2-2x^3)/(x(lnx^2)^2)$

What is the derivative of f(x) = (x^3-(lnx)^2)/(lnx^2)f(x)=x3−(lnx)2lnx2f(x) = (x^3-(lnx)^2)/(lnx^2)?