If `y=(e^x-e^(-x))/(e^x+e^(-x))` , prove that `(dy)/(dx)=1-y^2`

Explanation

We have, `{y}=frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}`

`frac{d y}{d x}=frac{d}{d x}(frac{e^{x}-e^{-x}}{e^{x}+e^{-x}})`

`=[frac{(e^{x}+e^{-x}) frac{d}{d x}(e^{x}-e^{-x})-(e^{x}-e^{-x}) frac{d}{d x}(e^{x}+e^{-x})}{(e^{x}+e^{-x})^{2}}]] `

`=[frac{(e^{x}+e^{-x}){e^{x}-e^{-x} frac{d}{d x}(-x)}-(e^{x}-e^{-x}){e^{x}+e^{-x} frac{d}{d x}(-x)}}{(e^{x}+e^{-x})^{2}}] `

`=[frac{(e^{x}+e^{-x})(e^{x}+e^{-x})-(e^{x}-e^{-x})(e^{x}-e^{-x})}{(e^{x}+e^{-x})^{2}}] `

`=frac{(e^{x}+e^{-x})^{2}-(e^{x}-e^{-x})^{2}}{(e^{x}+e^{-x})^{2}}] `

`=1-frac{(e^{x}-e^{-x})^{2}}{(e^{x}+e^{-x})^{2}} `

`=1-(frac{e^{x}-e^{-x}}{e^{x}+e^{-x}})^{2} `

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