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Differentials of inverse circular functions

We now turn to the inverse circular functions — arcsin, arccos, arctan, arcsec, arccosec, and arccot — and find their differentials using implicit differentiation.

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Take your time with each video. Pause, rewind, and think before moving on!

Dr Brian Brooks
Mathematics InSight

If \(\displaystyle{y=\arcsin x}\), what is \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}}\) in terms of \(\displaystyle{y}\) ?

\[x=\sin y\Rightarrow\, \frac{\mathrm{d}x}{\mathrm{d}y}=\cos y\]

What is \(\displaystyle{\cos y}\) in terms of \(x\) ?

\[\sin^2y+\cos^2y=1\]
\[\Rightarrow\, \cos^2y=1-x^2\]
\[\Rightarrow\, \cos y=\pm\sqrt{1-x^2}\]

Is \(\displaystyle{\cos y}\) positive or negative when \(y=\arcsin x\)?

\[-\frac{\pi}{2}\leq \arcsin x\leq \frac{\pi}{2}\Rightarrow\,-\frac{\pi}{2}\leq y\leq \frac{\pi}{2}\]
\[\Rightarrow\, \cos y\geq 0\]
\[\Rightarrow\, \cos y=\sqrt{1-x^2}\]

What are \(\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}y}\text{ and }\frac{\mathrm{d}y}{\mathrm{d}x}}\) in terms of \(\displaystyle{x}\) ?

\[\frac{\mathrm{d}x}{\mathrm{d}y}=\cos y=\sqrt{1-x^2}\]
\[\Rightarrow\, \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-x^2}}\]
unit circle for differentiation
\[\text{As }\delta\theta\to 0,\; \alpha\to\theta\]
\[\text{and }\delta x\to\sin\theta\times\delta\theta,\; \delta y\to\cos\theta\times\delta\theta\]
\[\Rightarrow\, \frac{\delta \theta}{\delta y}\to \frac{1}{\cos\theta}\]
\[=\frac{1}{\sqrt{1-y^2}}\text{ and }\theta=\arcsin y\]
\[\Rightarrow\, \frac{\mathrm{d}}{\mathrm{d}y}\arcsin y=\frac{1}{\sqrt{1-y^2}}\]
Find the differential of \(\arccos x\).
\[\begin{aligned}y&=\arccos x\Rightarrow x=\cos y\end{aligned}\]
\[\begin{aligned}\Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}&=-\sin y\end{aligned}\]
\[\begin{aligned}\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{1}{\sin y}=\pm\frac{1}{\sqrt{1-x^2}}\end{aligned}\]
\[\begin{aligned}\text{but gradient is always negative, and}\end{aligned}\]
\[\begin{aligned}0\le \arccos x=y\le \pi\Rightarrow \sin y\ge 0\end{aligned}\]
\[\begin{aligned}\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{1}{\sqrt{1-x^2}}\end{aligned}\]
Find the differential of \(\arctan x\).
\[\begin{aligned}y&=\arctan x\Rightarrow x=\tan y\end{aligned}\]
\[\begin{aligned}\Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}&=\sec^2 y\end{aligned}\]
\[\begin{aligned}\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{\sec^2 y}=\frac{1}{1+x^2}\end{aligned}\]

Well done!

You've now found the differentials of all six inverse circular functions: arcsin, arccos, arctan, arcsec, arccosec, and arccot.

What's Next:

  • These results open the door to a whole new class of integrals
  • Any integral whose answer involves an inverse circular function can now be recognised and evaluated

Dr Brian Brooks
Mathematics InSight

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