Question

e^Y-e^-y=2x prove that dy/dx= 1/√1-x^2
answer ​

Answer 1

Proof :

Given, $\displaystyle \mathsf{e^{y}-e^{-y}=2x}$ ...(i)

Squaring both sides, we get

$\displaystyle \mathsf{(e^{y}-e^{-y})^{2}=(2x)^{2}}$

$\displaystyle \to \mathsf{(e^{y}+e^{-y})^{2}-(4*e^{y}*e^{-y})=4x^{2}}$

$\displaystyle \to \mathsf{(e^{y}+e^{-y})^{2}=4(1+x^{2})}$

$\displaystyle \to \mathsf{e^{y}+e^{-y}=2\sqrt{1+x^{2}}}$

Differentiating both sides of (i) with respect to x, we get

$\displaystyle \mathsf{(e^{y}+e^{-y})\frac{dy}{dx}=2}$

$\displaystyle \to \mathsf{2\sqrt{1+x^{2}}\frac{dy}{dx}=2}$

$\displaystyle \to \mathsf{\frac{dy}{dx}=\frac{1}{\sqrt{1+x^{2}}}}$

Thus, proved.