Question

Find the second derivative of the following function:
y=log(base2)x/x²
y'=[(x/ln2)-2xlog(base2)x]/x^4
the derivative of x/ln2 is (ln2)^-1?
y"={[((ln2)^(-1))-(2log(base2)x/xln2)]x^4-4x³[(x/ln2)-2xlog(base2)x]}/x^8

I tried simplifying it and I got,
y"=[3/(x^4)ln2]-[2long(base2)x/(x^5)ln2]+[8log(base2)x/x^4]

Is the answer really off?

Thanks.
what a messy quesstion!

In your y'' , your forgot to treat the 2xlog<sub>2</sub>x as a product rule

I had
y' = [ x^4(1/ln2 - (2x(1/xln2) + 2log<sub>2</sub>x) - 4x^3(x/ln2 - 2xlog<sub>2</sub>x ]/x^8
= x^3[ x/ln2 - 2x/ln2 + 2xlog<sub>2</sub>x - 4/ln2 + 8xlog<sub>2</sub>x ]/x^8

= (10xlog<sub>2</sub>x - 5x/ln2)/x^5

check my steps to see if I am close.
Shouldn't it be -2xlog(base2)x because the product rule is inside the bracket, and its subtracting the bracket.

y"=[x/ln - 2xlog(base2)x -2x²/xln2 - 4x/ln2 + 8xlog(base2)x]/x^5

=[-3x/x^5] + [6xlog(base2)x/x^5] - [2x²/xln2]

Then I put them in common denominator,
[-5+6(log(base2)x)(ln2)]/(x^4)(ln2).

Thanks for guiding me!

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