Differentiate sin^-1 x

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

\(\frac{\mathrm{d} }{\mathrm{d} x} (\sin^{-1}x)= \frac{1}{\sqrt{1-x^{2}}}\)

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