If y=e^ x +e^ -x , prvoe that dy dx = y^2-4 - Vidyadip

Question

If $\text{y}=\text{e}^{\text{x}}+\text{e}^{-\text{x}},$ prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$

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Answer

Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}+\text{e}^{-\text{x}}\big)$
$=\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}+\frac{\text{d}}{\text{dx}}{\text{e}}^{-\text{x}}$
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}\big(-\text{x}\big)$
[Using chain rule]
$=\text{e}^{\text{x}}+\text{e}^{-\text{x}}(-1)$
$=\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)$
$=\sqrt{\big(\text{e}^{\text{x}}-\text{e}^{-\text{x}}\big)^2-4\text{e}^{\text{x}}\times\text{e}^{-\text{x}}}$
$\Big[\text{Since},(\text{a}-\text{b}=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}\Big]$
$=\sqrt{\text{y}^2-4}$
$\big[\text{Since e}^\text{x}+\text{e}^{-\text{x}}=\text{y}\big]$
Hence, the solution is, $\frac{\text{dy}}{\text{dx}}=\sqrt{\text{y}^2-4}$

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