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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation1 Mark
MCQ
Differential coefficient of $\sec(\tan^{-1}\text{x})$ is:
  • A
    $\frac{\text{x}}{1+\text{x}^2}$
  • B
    $\text{x}\sqrt{1+\text{x}^2}$
  • C
    $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
  • ✓
    $\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
✓

Answer

Correct option: D.
$\frac{\text{x}}{\sqrt{1+\text{x}^2}}$
We have, $\text{y}=\sec(\tan^{-1}\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sec\big(\tan^{-1}\text{x}\big)\tan\big(\tan^{-1}\text{x}\big)\times\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\sec\big(\tan^{-1}\text{x}\big)\tan\big(\tan^{-1}\text{x}\big)\times\frac{1}{\sqrt{1+\text{x}^2}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}\Big)\text{y}$

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