## - Differentiation of Inverse Trigonometric Functions

Here is a look at a formal proof for the derivations of the derivatives of $$\sin^{-1}⁡x, \cos^{-1}⁡x$$ and $$\tan^{-1}⁡x$$.

LESSON 1
Differentiate $$\sin^{-1}⁡x$$.
PROOF
Let $$y=\sin^{-1}⁡x$$ then $$\displaystyle \sin ⁡y=x, –{π\over 2}<y<{π\over2}$$
• Differentiate implicitly.
$$\displaystyle {d\over dx} (\sin ⁡y)={d\over dx} (x)$$
$$\displaystyle \cos ⁡y {dy\over dx}=1$$
• Solve for $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}={1\over \cos ⁡y}$$
RECALL: $$\cos^2⁡y+\sin^2⁡y=1$$
$$\cos^2⁡y=1-\sin^2⁡y$$
$$\cos⁡ y=\sqrt{1-\sin^2⁡y}=\sqrt{1-x^2}$$
• Rewrite $$\displaystyle {dy\over dx}={1\over \cos ⁡y}$$ in terms of $$x$$
$$\displaystyle {dy\over dx}={1\over \sqrt{1-x^2}}$$
$$\displaystyle {d\over dx} (\sin^{-1}⁡x)={1\over \sqrt{1-x^2}}$$

LESSON 2
Differentiate $$\cos^{-1}⁡x$$
PROOF
Let $$y=\cos^{-1}⁡x$$ then $$x=\cos ⁡y, 0\leq y\leq π$$
• Differentiate implicitly.
$$\displaystyle {d\over dx} (\cos⁡ y)={d\over dx}(x)$$
$$\displaystyle -\sin⁡ y {dy\over dx}=1$$
• Solve for $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}=-{1\over \sin ⁡y}$$
RECALL: $$\cos^2⁡y+\sin^2⁡y=1$$
$$\sin^2⁡y=1-\cos^2⁡y$$
$$\sin ⁡y=\sqrt{1-\cos^2⁡y}=\sqrt{1-x^2}$$
• Rewrite $$\displaystyle {dy\over dx}=-{1\over \sin ⁡y}$$ in terms of $$x$$.
$$\displaystyle {d\over dx} (\cos^{-1}⁡x)=-{1\over \sqrt{1-x^2}}$$

LESSON 3
Differentiate $$\tan^{-1}⁡x$$.
PROOF
Let $$y=\tan^{-1}⁡x$$ then $$\tan ⁡y=x, -{π\over 2}<y<{π\over2}$$
• Differentiate implicitly
$$\displaystyle {d\over dx} (\tan ⁡y)={d\over dx}(x)$$
$$\displaystyle \sec^2⁡ y {dy\over dx}=1$$
• Solve for $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}={1\over \sec^2⁡y}$$
RECALL: $$\sec^2⁡y=1+\tan^2⁡y=1+x^2$$
$$\displaystyle {d\over dx} (\tan^{-1}⁡x)={1\over 1+x^2}$$

LESSON 4
Show that if $$y=\cos^{-1}⁡3x$$, then $$\displaystyle y'=-{3\over \sqrt{1-9x^2}}$$
SOLUTION
$$y=\cos^{-1}⁡3x$$ iff $$3x=\cos ⁡y$$
• Differentiate implicitly
$$\displaystyle {d\over dx} (\cos ⁡y)={d\over dx}(3x)$$
$$\displaystyle -\sin ⁡y {dy\over dx}=3$$
• Solve for $$\displaystyle {dy\over dx}$$
$$\displaystyle {dy\over dx}=-{3\over \sin ⁡y}$$
RECALL: $$\sin ⁡y=\sqrt{1-\cos^2⁡y}=\sqrt{1-(3x)^2}$$
• Rewrite $$\displaystyle {dy\over dx}$$ in terms of $$x$$
$$\displaystyle {d\over dx} (\cos^{-1}⁡3x)=-{3\sqrt{1-(3x)^2}}$$
$$\displaystyle =-{3\over \sqrt{1-9x^2}}$$
In general, given that $$u$$ is a function of $$x$$, we have
$$\displaystyle y=\sin^{-1}⁡u \longrightarrow y'={1\over \sqrt{1-u^2}}u'$$
$$\displaystyle y=\cos^{-1}⁡u \longrightarrow y'=-{1\over \sqrt{1-u^2}}u'$$
$$\displaystyle y=\tan^{-1}⁡u \longrightarrow y'={1\over 1+u^2} u'$$