Calculus – Differentiation – The derivatives of the inverse trigonometric functions

Using implicit differentiation we are able to find the derivatives of the inverse trigonometric functions.

\(y=\arcsin(x)\;\Longleftrightarrow\;\sin(y)=x\) with \(-\frac{1}{2}\pi\leq y\leq\frac{1}{2}\pi\). Then we have: \(\cos(y)\cdot\displaystyle\frac{dy}{dx}=1\;\Longrightarrow\;\frac{dy}{dx}=\frac{1}{\cos(y)}\).

Now we have: \(\cos^2(y)=1-\sin^2(y)=1-x^2\;\Longrightarrow\;\cos(y)=\pm\sqrt{1-x^2}\).

Since \(-\frac{1}{2}\pi\leq y\leq\frac{1}{2}\pi\) it follows that \(\cos(y)\geq 0\) and therefore: \(\cos(y)=\sqrt{1-x^2}\). Conclusion: \(\displaystyle\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\).

\(y=\arccos(x)\;\Longleftrightarrow\;\cos(y)=x\) with \(0\leq y\leq\pi\). Then we have: \(-\sin(y)\cdot\displaystyle\frac{dy}{dx}=1\;\Longrightarrow\;\frac{dy}{dx}=-\frac{1}{\sin(y)}\).

Now we have: \(\sin^2(y)=1-\cos^2(y)=1-x^2\;\Longrightarrow\;\sin(y)=\pm\sqrt{1-x^2}\).

Since \(0\leq y\leq\pi\) it follows that \(\sin(y)\geq 0\) and therefore: \(\sin(y)=\sqrt{1-x^2}\). Conclusion: \(\displaystyle\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\).

\(y=\arctan(x)\;\Longleftrightarrow\;\tan(y)=x\) with \(-\frac{1}{2}\pi < y < \frac{1}{2}\pi\). Then we have: \(\displaystyle\frac{1}{\cos^2(y)}\cdot\frac{dy}{dx}=1\;\Longrightarrow\;\frac{dy}{dx}=\cos^2(y)\).

Now we have: \(\displaystyle\tan^2(y)=\frac{\sin^2(y)}{\cos^2(y)}=\frac{1-\cos^2(y)}{\cos^2(y)}\;\Longrightarrow\;\frac{1}{\cos^2(y)}=1+\tan^2(y)=1+x^2 \;\Longrightarrow\;\cos^2(y)=\frac{1}{1+x^2}\).

Conclusion: \(\displaystyle\frac{dy}{dx}=\frac{1}{1+x^2}\).

Résumé:

\begin{align*}\frac{d}{dx}(\arcsin(x))&=\frac{1}{\sqrt{1-x^2}}\\[5mm] \frac{d}{dx}(\arccos(x))&=-\frac{1}{\sqrt{1-x^2}}\\[5mm] \frac{d}{dx}(\arctan(x))&=\frac{1}{1+x^2}\end{align*}

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Last modified on March 1, 2021