Mathematics - Differentiation Question with Solution | TestHub

Let $y=\frac{x^2-x}{x^2+2x}$ $$\begin{array}{rl}& \Rightarrow \left(x^2+2x\right)y=x^2-x\\ & \Rightarrow x(x+2)y=x(x-1)\\ & \Rightarrow x[(x+2)y-(x-1)]=0\\ & \because x\ne 0,\therefore (x+2)y-(x-1)=0\\ & \Rightarrow xy+2y-x+1=0\\ & \Rightarrow x(y-1)=-(2y+1)\\ & \therefore x=\frac{2y+1}{1-y}\Rightarrow f^{-1}(x)=\frac{2x+1}{1-x}\\ & \frac{d}{dx}\left(f^{-1}(x)\right)=\frac{2(1-x)-(2x+1)(-1)}{(1-x{)}^2}\\ & =\frac{2-2x+2x+1}{(1-x{)}^2}=\frac{3}{(1-x{)}^2}\end{array}$$