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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

\(\begin{array}{l}\sin^{-1}x\end{array} \)

Let us now find the derivative of Inverse trigonometric function

Example: Find the derivative of a function

\(\begin{array}{l}y = \sin^{-1}x\end{array} \)
.

Solution:Given

\(\begin{array}{l}y = \sin^{-1}x\end{array} \)
…………(i)

\(\begin{array}{l}\Rightarrow x = \sin y\end{array} \)

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

\(\begin{array}{l}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\cos y}\end{array} \)

Putting the value of y form (i), we get

\(\begin{array}{l}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{\cos y} = \frac{1}{\cos (\sin^{-1}x)}\end{array} \)
………..(ii)

From equation (ii), we can see that the value of cos y cannot be equal to 0, as the function would become undefined.

\(\begin{array}{l}\Rightarrow \sin^{-1}x \neq \frac{-\pi}{2}, \frac{\pi}{2}\end{array} \)

i.e.

\(\begin{array}{l}x \neq -1,1\end{array} \)

From (i) we have

\(\begin{array}{l}y = \sin^{-1}x\end{array} \)

\(\begin{array}{l}\Rightarrow \sin y = \sin (\sin^{-1}x)\end{array} \)

Using property of trigonometric function,

\(\begin{array}{l}\cos^{2}y = 1 – \sin^{2}y = 1 – (\sin (\sin^{-1}x))^{2} = 1 – x^{2}\end{array} \)

\(\begin{array}{l}\Rightarrow \cos y = \sqrt{1 – x^{2}}\end{array} \)
…………(iii)

Now putting the value of (iii) in (ii), we have

\(\begin{array}{l}\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{1}{\sqrt{1-x^{2}}}\end{array} \)

Therefore, the Derivative of Inverse sine function is

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

Derivatives of Inverse trigonometric functions

Function (
\(\begin{array}{l}\frac{\mathrm{d} y}{\mathrm{d} x}\end{array} \)
)
arcsin x
\(\begin{array}{l}\frac{1}{\sqrt{1-x^{2}}}\end{array} \)
arccos x
\(\begin{array}{l}-\frac{1}{\sqrt{1-x^{2}}}\end{array} \)
arctan x
\(\begin{array}{l}\frac{1}{1+ x^{2}}\end{array} \)
arccot x
\(\begin{array}{l}-\frac{1}{1+ x^{2}}\end{array} \)
arcsec x
\(\begin{array}{l}\frac{1}{\left | x \right |\sqrt{x^{2}-1}}\end{array} \)
arccsc x
\(\begin{array}{l}-\frac{1}{\left | x \right |\sqrt{x^{2}-1}}\end{array} \)
Example:Find the derivative of a function 2 arcsin x – 5 arccsc x.

Solution:Given

\(\begin{array}{l}2\; arcsin x – 5\; arccsc x\end{array} \)

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

Further we can factorize the given expression.

Example:Find the derivative of a function

\(\begin{array}{l}\sin^{-1} \left ( \frac{1-x^{2}}{1+x^{2}} \right )\end{array} \)
.

Solution:Given y =

\(\begin{array}{l}\sin^{-1} \left ( \frac{1-x^{2}}{1+x^{2}} \right )\end{array} \)

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

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

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

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

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

\(\begin{array}{l}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-2}{1+x^{2}}\end{array} \)

Video Lesson on Trigonometry

Quiz on Derivative of Inverse Trigonometric functions

1 Comment

  1. Nice explain 😃 thank you so much for all byjus Teachers 🙏🖤

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