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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $y=\cos ^{-1}\left(e^x\right)$, then $\frac{d y}{d x}$ is :
  • A
    $\frac{1}{\sqrt{e^{-2 x}+1}}$
  • B
    $-\frac{1}{\sqrt{e^{-2 x}+1}}$
  • C
    $-\frac{1}{\sqrt{e^{-2 x}-1}}$
  • D
    $-\frac{1}{\sqrt{e^{-2 x}-1}}$
✓

Answer

We have, $y=\cos ^{-1}\left(e^x\right)$
\[\Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(e^x\right)^2}} e^x=\frac{-e^x}{\sqrt{1-e^{2 x}}}=\frac{-e^x}{e^x \sqrt{e^{-2 x}-1}}=\frac{-1}{\sqrt{e^{-2 x}-1}}\]

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