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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
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Answer

Let, $\text{y}=\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big]$
$=\Bigg[\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\frac{\text{d}}{\text{dx}}\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{x}}-\text{e}^{4\text{x}}\big)^2}\Bigg]$
[Using quotient rule and chain rule]
$=\frac{(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}})\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})+\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\Big[\text{e}^{2\text{x}}\frac{\text{d}}{\text{dx}}(2\text{x})-\text{e}^{-2\text{x}}\frac{\text{d}}{\text{dx}}(-2\text{x})\Big]}{\big(\text{e}^{2\text{x}}-\text{e}^{2\text{x}}\big)^2}$
$=\frac{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{x}}-2\text{e}^{-2\text{e}}\big)-\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)\big(2\text{e}^{2\text{e}}+2\text{e}^{-2\text{x}}\big)}{\big(\text{e}^{2\text{e}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2-2\big(\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}\big)^2}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{2\big[\text{e}^{4\text{x}}+\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}-\text{e}^{4\text{x}}-\text{e}^{-4\text{x}}-2\text{e}^{2\text{x}}\text{e}^{-2\text{x}}\big]}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
$=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^{2\text{x}}+\text{e}^{-2\text{x}}}{\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}}\Big)=\frac{-8}{\big(\text{e}^{2\text{x}}-\text{e}^{-2\text{x}}\big)^2}$

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