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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 \(\displaystyle{y}\) ?

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

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

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

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

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

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

\[\frac{\mathrm{d}x}{\mathrm{d}y}=\cos y=\sqrt{1-x^2}\]
\[\implies\, \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-x^2}}\]
Find the differential of \(\arccos x\).
\[\begin{aligned}y&=\arccos x\implies x=\cos y\end{aligned}\]
\[\begin{aligned}\implies \frac{\mathrm{d}x}{\mathrm{d}y}&=-\sin y\end{aligned}\]
\[\begin{aligned}\implies \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{1}{\sin y}=\pm\frac{1}{\sqrt{1-x^2}}\end{aligned}\]
\[\begin{aligned}\text{but gradient is always negative, and}\end{aligned}\]
\[\begin{aligned}0\le \arccos x=y\le \pi\implies \sin y\ge 0\end{aligned}\]
\[\begin{aligned}\implies \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{1}{\sqrt{1-x^2}}\end{aligned}\]
Find the differential of \(\arctan x\).
\[\begin{aligned}y&=\arctan x\implies x=\tan y\end{aligned}\]
\[\begin{aligned}\implies \frac{\mathrm{d}x}{\mathrm{d}y}&=\sec^2 y\end{aligned}\]
\[\begin{aligned}\implies \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{\sec^2 y}=\frac{1}{1+x^2}\end{aligned}\]
\begin{align*} \arcsin a-\arcsin b&=\arcsin\left(a\sqrt{1-b^2}-b\sqrt{1-a^2}\right)\\[15pt] \implies \frac{1}{h}\left(\arcsin (x+h)-\arcsin x\right)&=\arcsin\left((x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}\right)\\[15pt] &=\frac{1}{h}\arcsin\left(x\left(\sqrt{1-x^2}-\sqrt{1-(x+h)^2}\right)+h\sqrt{1-x^2}\right)\\[15pt] &=\frac{1}{h}\arcsin\left(x\frac{(1-x^2)-\left(1-(x+h)^2\right)}{\sqrt{1-x^2}+\sqrt{1-(x+h)^2}}+h\sqrt{1-x^2}\right)\\[15pt] &=\frac{1}{h}\arcsin\left(x\frac{(x+h)^2-x^2}{\sqrt{1-x^2}+\sqrt{1-(x+h)^2}}+h\sqrt{1-x^2}\right)\\[15pt] &=\frac{1}{h}\arcsin\left(x\frac{2xh+h^2}{\sqrt{1-x^2}+\sqrt{1-(x+h)^2}}+h\sqrt{1-x^2}\right)\\[15pt] &\to \lim_{h\to0}\frac{1}{h}\arcsin \left( x\frac{2xh}{2\sqrt{1-x^2}}+h\sqrt{1-x^2}\right)\\[15pt] &= \lim_{h\to0}\frac{1}{h}\arcsin\frac{h(x^2+(1-x^2))}{\sqrt{1-x^2}}\\[15pt] &= \lim_{h\to0}\frac{1}{h}\arcsin\frac{h}{\sqrt{1-x^2}}\\[15pt] &= \frac{1}{\sqrt{1-x^2}}\lim_{h\to0}\frac{\sqrt{1-x^2}}{h}\arcsin\frac{h}{\sqrt{1-x^2}}\\[15pt] &= \frac{1}{\sqrt{1-x^2}}\lim_{\theta\to0}\frac{1}{\theta}\arcsin\theta\\[15pt] &= \frac{1}{\sqrt{1-x^2}} \end{align*}
\begin{align*} \text{Let }\varphi&=\arccos(x+h),\quad\theta=\arccos x\quad(\text{so }\cos\varphi=x+h,\ \cos\theta=x)\\[15pt] \implies \cos\varphi-\cos\theta&=h\\[15pt] \implies -2\sin\left(\frac{\varphi+\theta}{2}\right)\sin\left(\frac{\varphi-\theta}{2}\right)&=h\\[15pt] \implies \frac{\varphi-\theta}{h}&=-\frac{1}{\sin\left(\dfrac{\varphi+\theta}{2}\right)}\cdot\frac{\left(\varphi-\theta\right)/2}{\sin\left(\dfrac{\varphi-\theta}{2}\right)}\\[15pt] &\to -\frac{1}{\sin\theta}\lim_{u\to0}\frac{u}{\sin u}\quad\left(h\to0\implies\varphi\to\theta,\ u=\frac{\varphi-\theta}{2}\to0\right)\\[15pt] &=-\frac{1}{\sin\theta}\\[15pt] \implies \frac{\mathrm{d}}{\mathrm{d}x}\arccos x&=-\frac{1}{\sin\theta}=-\frac{1}{\sqrt{1-x^2}}\quad(\sin\theta\ge0\text{ since }\theta\in[0,\pi]) \end{align*}
\begin{align*} \frac{\arctan(x+h)-\arctan x}{h}&=\frac{1}{h}\arctan\frac{x+h-x}{1+x(x+h)}\\[15pt] &=\frac{1}{h}\arctan\frac{h}{1+x(x+h)}\\[15pt] &=\frac{\left(1+x(x+h)\right)}{h\left(1+x(x+h)\right)}\arctan\frac{h}{1+x(x+h)}\\[15pt] &=\frac{1}{\left(1+x(x+h)\right)}\frac{\arctan\theta}{\theta}\text{ where }\theta=\frac{h}{1+x(x+h)}\\[15pt] &\to \frac{1}{1+x^2}\lim_{\theta\to 0} \frac{\arctan\theta}{\theta}\\[15pt] &=\frac{1}{1+x^2} \end{align*}
unit circle for differentiation

Although I know that the circular functions are, rather obviously, related to circles, I was still intrigued by the fact that the differentials of the inverse functions look so much like the equation of a circle.

To satisfy my curiosity, I went back to a diagram from earlier in the Differentials Extra sheet. Here is a summary of my thinking — it's not a detailed walk-through, more of an expository excursion.

First, notice that \(\delta x < 0\), even though it is attached in the diagram to a positive distance.

\[\text{As }\delta\theta\to 0,\; \alpha\to\theta\]
\[\text{and }\delta x\to-\sin\theta\times\delta\theta,\quad \delta y\to\cos\theta\times\delta\theta\]
\[\implies\, \frac{\delta \theta}{\delta y}\to \frac{1}{\cos\theta}\qquad \frac{\delta \theta}{\delta x}\to -\frac{1}{\sin\theta}\]
\[\implies\, \frac{\mathrm{d} \theta}{\mathrm{d} y}= \frac{1}{\cos\theta}\qquad \frac{\mathrm{d} \theta}{\mathrm{d} x}= -\frac{1}{\sin\theta}\]
\[\text{ and }\theta=\arcsin y=\arccos x\]
\[\implies\, \frac{\mathrm{d}}{\mathrm{d}y}\arcsin y=\frac{1}{\cos\theta}=\frac{1}{\sqrt{1-y^2}}\]
\[\frac{\mathrm{d}}{\mathrm{d}x}\arccos x=-\frac{1}{\sin\theta}=-\frac{1}{\sqrt{1-x^2}}\]

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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