Question

If \[y = e^x + e^{- x}\] prove that  \[\frac{dy}{dx} = \sqrt{y^2 - 4}\] ?

Answer 1

\[\text{We have, }y = e^x + e^{- x}\]

Differentiate it with respect to x, 

\[\frac{d y}{d x} = \frac{d}{dx}\left( e^x + e^{- x} \right)\]

\[ = \frac{d}{dx} e^x + \frac{d}{dx} e^{- x} \]

\[ = e^x + e^{- x} \frac{d}{dx}\left( - x \right) .........\left[ \text{Using chain rule } \right]\]

\[ = e^x + e^{- x} \left( - 1 \right)\]

\[ = \left( e^x - e^{- x} \right)\]

\[ = \sqrt{\left( e^x + e^{- x} \right)^2 - 4 e^x \times e^{- x}} ..........\left[ \because \left( a - b \right) = \sqrt{\left( a + b \right)^2 - 4ab} \right]\]

\[ = \sqrt{y^2 - 4} ..........\left [ \because e^x + e^{- x} = y \right]\]