Differentiation of log x w.r.t 1/x is :

The correct answer is Option (3) → $-x$

$\frac{d(\log x)}{dz}=\frac{1}{x}\frac{dx}{dz}$

$\frac{d(\frac{1}{x})}{dz}=-\frac{1}{x^2}\frac{dx}{dz}$

$∴\frac{d(\log x)}{d(\frac{1}{-x})}=\frac{\frac{1}{x}}{\frac{1}{x^2}}=\frac{-x^2}{x}=-x$