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

Jun 24, 2018

$f ' \left(x\right) = \frac{1}{2 \sqrt{\ln {x}^{2} - x}} \left(\frac{2}{x} - 1\right)$

Explanation:

We will differentiate $f$ using the chain rule:
$y = f \left(g \left(x\right)\right) \implies y ' = f ' \left(g \left(x\right)\right) g ' \left(x\right)$
Therefore, $f ' \left(x\right) = \frac{1}{2 \sqrt{\ln \left({x}^{2}\right) - x}} \frac{d}{\mathrm{dx}} \left(\ln {x}^{2} - x\right)$
To differentiate $\ln {x}^{2} - x$, recognize that $\ln {x}^{2} = 2 \ln x$ so $\frac{d}{\mathrm{dx}} \left(\ln {x}^{2} - x\right) = \frac{2}{x} - 1$. Therefore, $f ' \left(x\right) = \frac{1}{2 \sqrt{\ln {x}^{2} - x}} \left(\frac{2}{x} - 1\right)$.

Given: $f \left(x\right) = \setminus \sqrt{\setminus \ln \left({x}^{2}\right) - x}$

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

$\setminus \frac{d}{\mathrm{dx}} f \left(x\right) = \setminus \frac{d}{\mathrm{dx}} \setminus \sqrt{\setminus \ln \left({x}^{2}\right) - x}$

$f ' \left(x\right) = \setminus \frac{1}{2 \setminus \sqrt{\setminus \ln \left({x}^{2}\right) - x}} \setminus \frac{d}{\mathrm{dx}} \left(\setminus \ln \left({x}^{2}\right) - x\right)$

$f ' \left(x\right) = \setminus \frac{1}{2 \setminus \sqrt{\setminus \ln \left({x}^{2}\right) - x}} \left(\setminus \frac{1}{{x}^{2}} \setminus \cdot 2 x - 1\right)$

$f ' \left(x\right) = \setminus \frac{2 - x}{2 x \setminus \sqrt{\setminus \ln \left({x}^{2}\right) - x}}$