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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
✓

Answer

Let $\text{y}=\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
Differentiate with respect to x we get,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}\log\text{x})-(\text{e}^\text{x}\log\text{x})\frac{\text{d}}{\text{dx}}\text{x}^2}{\big(\text{x}^2\big)^2}$
[Using quotient rule]
$=\frac{\text{x}^2\Big\{\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})\Big\}-\text{e}^\text{x}\log\text{x}\times2\text{x}}{\text{x}^4}$
[Using product rule]
$=\frac{\text{x}^2\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]-2\text{xe}^\text{x}\log\text{x}}{\text{x}^4}$
$=\frac{\frac{\text{x}^2\text{e}^\text{z}(1+\text{x}\log\text{x})}{\text{x}}-2\text{xe}^\text{z}\log\text{x}}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}[1+\text{x}\log\text{x}-2\log\text{x}]}{\text{x}^4}$
$=\frac{\text{xe}^\text{x}}{\text{x}^3}\Big[\frac{1}{\text{x}}+\frac{\text{x}\log\text{x}}{\text{x}}-\frac{2\log\text{x}}{\text{x}}\Big]$
$=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$
So,
$\frac{\text{d}}{\text{dx}}\Big[\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}\Big]=\text{e}^\text{x}\text{x}^{-2}\Big[\frac{1}{\text{x}}+\log\text{x}-\frac{2}{\text{x}}\log\text{x}\Big]$

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