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Question:

If $y=e^{x}+e^{-x}$, prove that $\frac{d y}{d x}=\sqrt{y^{2}-4} .$

Solution:

Given $y=e^{x}+e^{-x}$

On differentiating $y$ with respect to $x$, we get

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

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(e^{x}\right)+\frac{d}{d x}\left(e^{-x}\right)$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}}$

$\Rightarrow \frac{d y}{d x}=e^{x}+e^{-x} \frac{d}{d x}(-x)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=e^{x}-e^{-x} \frac{d}{d x}(x)$

We have $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$

$\Rightarrow \frac{d y}{d x}=e^{x}-e^{-x} \times 1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\left(\mathrm{e}^{\mathrm{x}}\right)^{2}+\left(\mathrm{e}^{-\mathrm{x}}\right)^{2}-2\left(\mathrm{e}^{\mathrm{x}}\right)\left(\mathrm{e}^{-\mathrm{x}}\right)}$

$\Rightarrow \frac{d y}{d x}=\sqrt{\left(e^{x}\right)^{2}+\left(e^{-x}\right)^{2}-2\left(e^{x}\right)\left(e^{-x}\right)+2\left(e^{x}\right)\left(e^{-x}\right)-2\left(e^{x}\right)\left(e^{-x}\right)}$

$\Rightarrow \frac{d y}{d x}=\sqrt{\left(e^{x}\right)^{2}+\left(e^{-x}\right)^{2}+2\left(e^{x}\right)\left(e^{-x}\right)-4\left(e^{x}\right)\left(e^{-x}\right)}$

$\Rightarrow \frac{d y}{d x}=\sqrt{\left(e^{x}+e^{-x}\right)^{2}-4}$

But, $y=e^{x}+e^{-x}$

$\therefore \frac{d y}{d x}=\sqrt{y^{2}-4}$

Thus, $\frac{d y}{d x}=\sqrt{y^{2}-4}$

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