The inverse trigonometric function are represented by adding arc in prefix for a trigonometric function, or by adding the power of -1, such as:
Inverse of sin x = arcsin(x) or \(\sin^{-1}x\)
Given \(y = \sin^{-1}x\)…………(i)
\(\Rightarrow x = \sin y\)Differentiating the above equation w.r.t. x, we have:
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\cos y}\)Putting the value of y form (i), we get
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\cos y} = \frac{1}{\cos (\sin^{-1}x)}\)………..(ii)From equation (ii), we can see that the value of cos y cannot be equal to 0, as the function would become undefined.
\(\Rightarrow \sin^{-1}x \neq \frac{-\pi}{2}, \frac{\pi}{2}\) i.e. \(x \neq -1,1\)From (i) we have
\(y = \sin^{-1}x\) \(\Rightarrow \sin y = \sin (\sin^{-1}x)\)Using property of trigonometric function,
\(\cos^{2}y = 1 – \sin^{2}y = 1 – (\sin (\sin^{-1}x))^{2} = 1 – x^{2}\) \(\Rightarrow \cos y = \sqrt{1 – x^{2}}\)…………(iii)Now putting the value of (iii) in (ii), we have
\(\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\sqrt{1-x^{2}}}\)Therefore, the Derivative of Inverse sine function is
\(\frac{\mathrm{d} }{\mathrm{d} x} (\sin^{-1}x)= \frac{1}{\sqrt{1-x^{2}}}\)
