- 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
\(\displaystyle y=\sin^{-1}⁡u \longrightarrow y'={1\over \sqrt{1-u^2}}u'\)
\(\displaystyle y=\cos^{-1}⁡u \longrightarrow y'=-{1\over \sqrt{1-u^2}}u'\)
\(\displaystyle y=\tan^{-1}⁡u \longrightarrow y'={1\over 1+u^2} u'\)