Derivative of Inverse Trigonometric functions

The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. These functions are used to obtain angle for a given trigonometric value. Inverse trigonometric functions have various application in engineering, geometry, navigation etc.

Representation of functions:

Generally, 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 \(\sin^{-1}x\)

Let us now find the derivative of Inverse trigonometric function

Example: Find the derivative of a function \(y = \sin^{-1}x\).

Solution:Given \(y = \sin^{-1}x\)…………(i)

\(\Rightarrow x = \sin y\)

Differentiating the above equation w.r.t. x, we have

\(dx = \cos y. \frac{\mathrm{d} y}{\mathrm{d} x}\)

\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\cos y}\)

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. \(x \neq -1,1\)

\(\Rightarrow x = (-1,1)\)

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

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

Derivatives of Inverse trigonometric functions:

Function

(\(\frac{\mathrm{d} y}{\mathrm{d} x}\)

arcsin x

\(\frac{1}{\sqrt{1-x^{2}}}\)

arccos x

\(-\frac{1}{\sqrt{1-x^{2}}}\)

arctan x

\(\frac{1}{1+ x^{2}}\)

arccot x

\(-\frac{1}{1+ x^{2}}\)

arcsec x

\(\frac{1}{\left | x \right |\sqrt{x^{2}-1}}\)

arccsc x

\(-\frac{1}{\left | x \right |\sqrt{x^{2}-1}}\)

Example:Find the derivative of a function 2 arcsin x – 5 arccsc x.

Solution:Given \(2\; arcsin x – 5\; arccsc x\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{2}{\sqrt{1-x^{2}}} + \frac{5}{ x \sqrt{x^{2}-1}}\)

Further we can factorize the given expression.

Example:Find the derivative of a function \(\sin^{-1} \left ( \frac{1-x^{2}}{1+x^{2}} \right )\).

Solution:Given y = \(\sin^{-1} \left ( \frac{1-x^{2}}{1+x^{2}} \right )\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\sqrt{1-\left ( \frac{1-x^{2}}{1+x^{2}} \right )^{2}}} \times \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{1-x^{2}}{1+x^{2}} \right )\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\sqrt{ \frac{(1+x^{2})^{2}- (1-x^{2})^{2}}{(1+x^{2})^{2}}} } \times \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{1-x^{2}}{1+x^{2}} \right )\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1+x^{2}}{\sqrt{ (1+x^{4}+2x^{2})- (1+x^{4}-2x^{2})} } \times \left ( \frac{(1+x^{2})(-2x)- (1-x^{2})(2x))}{(1+x^{2})^{2}} \right )\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1+x^{2}}{\sqrt{ 4x^{2}} } \times \left ( \frac{(-2x -2x^{3}- 2x + 2x^{3}) }{(1+x^{2})^{2}} \right )\)

\(\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1+x^{2}}{2x} \times \left ( \frac{-4x }{(1+x^{2})^{2}} \right )\)

\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-2}{1+x^{2}}\)<


Practise This Question

If 2x+2y=2x+y, then dydx is equal to