Question

How do you differentiate `f(x)= sqrt (lnx^2-x)`?

Answer 1

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

Explanation:

We will differentiate `f` using the chain rule:
`y=f(g(x))=>y'=f'(g(x))g'(x)`
Therefore, `f'(x)=1/(2sqrt(ln(x^2)-x))d/dx (lnx^2-x)`
To differentiate `lnx^2-x`, recognize that `lnx^2=2lnx` so `d/dx (lnx^2-x)=2/x-1`. Therefore, `f'(x)=1/(2sqrt(lnx^2-x))(2/x-1)`.

Answer 2

Given: `f(x)=\sqrt{\ln(x^2)-x}`

Differentiating w.r.t. `x` as follows

`\frac{d}{dx}f(x)=\frac{d}{dx}\sqrt{\ln(x^2)-x}`

`f'(x)=\frac{1}{2\sqrt{\ln(x^2)-x}}\frac{d}{dx}(\ln(x^2)-x)`

`f'(x)=\frac{1}{2\sqrt{\ln(x^2)-x}}(\frac{1}{x^2}\cdot 2x-1)`

`f'(x)=\frac{2-x}{2x\sqrt{\ln(x^2)-x}}`