Question

Differentiate $x^2-3x+6$ w.r.t $\frac{1}{x}$

Answer 1

$f\left(x\right)=x^2-3x+6g\left(x\right)=\frac{1}{x}{f}^{\prime }\left(x\right)=\frac{d}{dx}{x}^2-3x+6=2x-3g^{\prime }\left(x\right)=\frac{d}{dx}\frac{1}{x}=\log x\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}=\frac{2x-3}{\log x}$