Question Number 204689 by Davidtim last updated on 25/Feb/24
$${y}=\mid{f}\left({x}\right)\mid\:\:;\:\:\:\:\frac{{dy}}{{dx}}=? \\ $$
Answered by A5T last updated on 25/Feb/24
![y=(√((f(x))^2 ))=[(f(x))^2 ]^(1/2) ⇒(dy/dx)=(1/(2(√((f(x))^2 ))))×2f(x)×f′(x)=((f(x)f^′ (x))/(∣f(x)∣)) =sgn(f(x))f′(x)](https://www.tinkutara.com/question/Q204692.png)
$${y}=\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} }=\left[\left({f}\left({x}\right)\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\left({f}\left({x}\right)\right)^{\mathrm{2}} }}×\mathrm{2}{f}\left({x}\right)×{f}'\left({x}\right)=\frac{{f}\left({x}\right){f}^{'} \left({x}\right)}{\mid{f}\left({x}\right)\mid} \\ $$$$={sgn}\left({f}\left({x}\right)\right){f}'\left({x}\right) \\ $$