Kara Cummings

2023-03-21

The differential coefficient of $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$.

alexccheng68b6a1

Beginner2023-03-22Added 9 answers

Determine the function's differential coefficient.

The function $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$ is equivalent to the expression:

$\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)=\mathrm{sec}\left({\mathrm{sec}}^{-1}\left({x}^{2},+,1\right)\right)\left(\because {\mathrm{tan}}^{-1}\left(x\right)={\mathrm{sec}}^{-1}\left({x}^{2},+,1\right)\right)={x}^{2}+1$

The function is a compound function.

Determine the derivative:

$\frac{d}{dx}\left({x}^{2},+,1\right)=\frac{1}{2{x}^{2}+1}\frac{d}{dx}\left({x}^{2}+1\right)\Rightarrow \frac{d}{dx}\left({x}^{2},+,1\right)=\frac{1}{2{x}^{2}+1}\left(2x\right)\Rightarrow \frac{d}{dx}\left({x}^{2},+,1\right)=\frac{x}{{x}^{2}}$

Hence, the differential coefficient is $\frac{x}{{x}^{2}}$.

The function $\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$ is equivalent to the expression:

$\mathrm{sec}\left({\mathrm{tan}}^{-1}\left(x\right)\right)=\mathrm{sec}\left({\mathrm{sec}}^{-1}\left({x}^{2},+,1\right)\right)\left(\because {\mathrm{tan}}^{-1}\left(x\right)={\mathrm{sec}}^{-1}\left({x}^{2},+,1\right)\right)={x}^{2}+1$

The function is a compound function.

Determine the derivative:

$\frac{d}{dx}\left({x}^{2},+,1\right)=\frac{1}{2{x}^{2}+1}\frac{d}{dx}\left({x}^{2}+1\right)\Rightarrow \frac{d}{dx}\left({x}^{2},+,1\right)=\frac{1}{2{x}^{2}+1}\left(2x\right)\Rightarrow \frac{d}{dx}\left({x}^{2},+,1\right)=\frac{x}{{x}^{2}}$

Hence, the differential coefficient is $\frac{x}{{x}^{2}}$.

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