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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x},$ then $\frac{\text{dy}}{\text{dx}}=$
  • ✓
    $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$
  • B
    $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$
  • C
    $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big\{\log(\text{x}+1)-\frac{\text{x}}{\text{x}+1}\Big\}$
  • D
    $\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big\{\log\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}\Big\}$
✓

Answer

Correct option: A.
$\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}+1}$
Let $\text{y}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}$
Taking log on both sides,
$\log\text{y}=\text{x}\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\bigg(\frac{1}{1+\frac{1}{\text{x}}}\bigg)\frac{\text{d}}{\text{dx}}\Big(1+\frac{1}{\text{x}}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\times\frac{\text{x}}{\text{x}+1}\Big(-\frac{1}{\text{x}^2}\Big)+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{x}+1}\times-\frac{1}{\text{x}^2}+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{-1}{\text{x}+1}+\log\Big(1+\frac{1}{\text{x}}\Big)\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(1+\frac{1}{\text{x}}\Big)^\text{x}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)-\frac{-1}{\text{x}+1}\Big]$

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