# find the derivative of the following function f(x)=log(x^2_1/2x)

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### 2 Answers

You need to change the base of logarithm such that:

`log_(x^2) (1/2x) = ln (1/2x)/ ln (x^2)`

You need to differentiate the function with respect to x, using the quotient rule:

`f(x) = ln (1/2x)/ ln (x^2) =gt f'(x) = {[ln (1/2x)]'*ln (x^2) - ln (1/2x)*[ ln (x^2)]'}/(ln (x^2))^2`

`f'(x) = ((((1/2x)')/(1/2x))*ln (x^2) - (2x/x^2)*(ln (1/2x)))/((ln (x^2)^2)`

`f'(x) = ((-ln (x^2))/(x) - (ln (1/2x)^2)/(x)) / (ln (x^2)^2)`

`f'(x) = (ln(1/x^2) - ln (1/2x)^2)/(x*(ln (x^2))^2)`

`f'(x) = (ln((1/x^2)/(1/4x^2)))/(x*(ln (x^2))^2)`

`f'(x) = ln 4/(x*(ln (x^2))^2)`

**Hence, the derivative of the function is `f'(x) = ln 4/(x*(ln (x^2))^2).` **

I'm not sure about the question you ask, but if by "_" you mean "-" as in subtraction, I think one possible solution is:

f'(x) = d(ln(x^2-1/2x)/ln2)/dx

As d(lnk)/dk = 1/k, let u = x^2-1/2x, we have:

f'(x) = d(ln(u)/ln2)/du * du/dx (chain rule)

f'(x) = u'(x) * 1/(u*ln2)

f'(x) = (2x-1/2)/[(x^2-1/2x)*ln2]

**The derivative of f(x)=log(x^2_1/2x)base 2 is **f'(x) = (2x-1/2)/[(x^2-1/2x)*ln2]