Question

If $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$, prove that $\frac{dy}{dx}=1-y^2$

Answer 1

$\mathrm{We}\mathrm{have},y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$
Differentiating with respect to x,
$$\begin{gathered} \frac{\mathit{d}y}{\mathit{d}x}=\frac{d}{dx}\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right) \\ =\left[\frac{\left(e^x+e^{-x}\right){\displaystyle \frac{d}{dx}}\left(e^x-e^{-x}\right)-\left(e^x-e^{-x}\right){\displaystyle \frac{d}{dx}}\left(e^x+e^{-x}\right)}{{\left(e^x+e^{-x}\right)}^2}\right] \\ =\left[\frac{\left(e^x+e^{-x}\right)\left\{e^x-e^{-x}{\displaystyle \frac{d}{dx}}\left(-x\right)\right\}-\left(e^x-e^{-x}\right)\left\{e^x+e^{-x}{\displaystyle \frac{d}{dx}}\left(-x\right)\right\}}{{\left(e^x+e^{-x}\right)}^2}\right] \\ =\left[\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{{\left(e^x+e^{-x}\right)}^2}\right] \\ =\frac{{\left(e^x+e^{-x}\right)}^2-{\left(e^x-e^{-x}\right)}^2}{{\left(e^x+e^{-x}\right)}^2} \\ =1-\frac{{\left(e^x-e^{-x}\right)}^2}{{\left(e^x+e^{-x}\right)}^2} \\ =1-{\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)}^2 \\ =1-y^2 \\ \mathrm{So},\frac{\mathit{d}y}{\mathit{d}x}=1-y^2 \end{gathered}$$