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 [latex]\sin^{-1}x[/latex]
Given [latex]y = \sin^{-1}x[/latex]…………(i)
[latex]\Rightarrow x = \sin y[/latex]Differentiating the above equation w.r.t. x, we have:
[latex]\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\cos y}[/latex]Putting the value of y form (i), we get
[latex]\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\cos y} = \frac{1}{\cos (\sin^{-1}x)}[/latex]………..(ii)From equation (ii), we can see that the value of cos y cannot be equal to 0, as the function would become undefined.
[latex]\Rightarrow \sin^{-1}x \neq \frac{-\pi}{2}, \frac{\pi}{2}[/latex] i.e. [latex]x \neq -1,1[/latex]From (i) we have
[latex]y = \sin^{-1}x[/latex] [latex]\Rightarrow \sin y = \sin (\sin^{-1}x)[/latex]Using property of trigonometric function,
[latex]\cos^{2}y = 1 – \sin^{2}y = 1 – (\sin (\sin^{-1}x))^{2} = 1 – x^{2}[/latex] [latex]\Rightarrow \cos y = \sqrt{1 – x^{2}}[/latex]…………(iii)Now putting the value of (iii) in (ii), we have
[latex]\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\sqrt{1-x^{2}}}[/latex]Therefore, the Derivative of Inverse sine function is
[latex]\frac{\mathrm{d} }{\mathrm{d} x} (\sin^{-1}x)= \frac{1}{\sqrt{1-x^{2}}}[/latex]
