rzepiowiffn

2023-03-14

What is the derivative of ${\mathrm{sec}}^{-1}\left(x\right)$?

kuskdtq

Beginner2023-03-15Added 4 answers

Let $y={\mathrm{sec}}^{-1}x$.

by rewriting in terms of secant,

$\Rightarrow \mathrm{sec}y=x$

by differentiating with respect to x,

$\Rightarrow \mathrm{sec}y\mathrm{tan}y\cdot y\prime =1$

by dividing by sec y tan y,

$\Rightarrow y\prime =\frac{1}{\mathrm{sec}y\mathrm{tan}y}$

since $\mathrm{sec}y=x$ and $\mathrm{tan}x=\sqrt{{\mathrm{sec}}^{2}y-1}=\sqrt{{x}^{2}-1}$

$\Rightarrow y\prime =\frac{1}{x\sqrt{{x}^{2}-1}}$

Thus,

$\frac{d}{dx}\left({\mathrm{sec}}^{-1}x\right)=\frac{1}{x\sqrt{{x}^{2}-1}}$

by rewriting in terms of secant,

$\Rightarrow \mathrm{sec}y=x$

by differentiating with respect to x,

$\Rightarrow \mathrm{sec}y\mathrm{tan}y\cdot y\prime =1$

by dividing by sec y tan y,

$\Rightarrow y\prime =\frac{1}{\mathrm{sec}y\mathrm{tan}y}$

since $\mathrm{sec}y=x$ and $\mathrm{tan}x=\sqrt{{\mathrm{sec}}^{2}y-1}=\sqrt{{x}^{2}-1}$

$\Rightarrow y\prime =\frac{1}{x\sqrt{{x}^{2}-1}}$

Thus,

$\frac{d}{dx}\left({\mathrm{sec}}^{-1}x\right)=\frac{1}{x\sqrt{{x}^{2}-1}}$

Veronella6nk

Beginner2023-03-16Added 6 answers

Answer: $\left({\mathrm{sec}}^{-1}\left(x\right)\right)\prime =\frac{1}{x\sqrt{{x}^{2}-1}}$

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