Question

What are the derivatives of the inverse trigonometric functions?

Answer 1

Hint: We need to find the derivative of the inverse trigonometric functions. We have direct formulae for the derivative of inverse trigonometric functions. To answer this question, we will list all the inverse trigonometric functions and their corresponding formulae for the derivative.

Complete step by step answer:
We denote the derivative of the function \[f\left( x \right)\] with respect to \[x\] as \[\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\].
Let us first list all the inverse trigonometric functions.
We have six inverse trigonometric functions.
The following are the six inverse trigonometric functions:
\[{\sin ^{ - 1}}x,{\cos ^{ - 1}}x,{\tan ^{ - 1}}x,{\csc ^{ - 1}}x,{\sec ^{ - 1}}x,{\cot ^{ - 1}}x\]
Now, let us write all these inverse trigonometric functions along with their derivatives.
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\], \[ - 1 < x < 1\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = - \dfrac{1}{{\sqrt {1 - {x^2}} }}\], \[ - 1 < x < 1\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}\], \[ - \infty < x < \infty \]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\csc }^{ - 1}}x} \right) = - \dfrac{1}{{|x|\sqrt {{x^2} - 1} }}\], \[x \in \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \dfrac{1}{{|x|\sqrt {{x^2} - 1} }}\], \[x \in \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {{{\cot }^{ - 1}}x} \right) = - \dfrac{1}{{1 + {x^2}}}\], \[ - \infty < x < \infty \]
Hence, the above are the formulae for the derivative of the inverse trigonometric functions.

Note:
We can also find the derivative of inverse trigonometric functions using the inverse function theorem. For example, let \[x = f\left( y \right) = \sin y\]is the inverse of the function \[y = g\left( x \right) = {\sin ^{ - 1}}x\]. Then the derivative of \[y = {\sin ^{ - 1}}x\] is given by \[g'\left( x \right) = \dfrac{1}{{f'\left( y \right)}} = \dfrac{1}{{\left( {\sin y} \right)'}}\], where \[f'\left( x \right) = \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)\].
Then, we know, \[\left( {\sin x} \right)' = \cos x\]. So, putting this, we get \[g'\left( x \right) = \dfrac{1}{{\cos y}}\]. Now, using \[{\cos ^2}x = 1 - {\sin ^2}x\], we get \[g'\left( x \right) = \dfrac{1}{{\sqrt {1 - {{\sin }^2}y} }} = \dfrac{1}{{\sqrt {1 - {{\sin }^2}\left( {{{\sin }^{ - 1}}x} \right)} }} = \dfrac{1}{{\sqrt {1 - {{\left( {\sin \left( {{{\sin }^{ - 1}}x} \right)} \right)}^2}} }}\]. Now, we know, \[\left( {\sin \left( {{{\sin }^{ - 1}}x} \right) = x} \right)\], so, we get, \[g'\left( x \right) = \dfrac{1}{{\sqrt {1 - {{\sin }^2}y} }} = \dfrac{1}{{\sqrt {1 - {{\sin }^2}\left( {{{\sin }^{ - 1}}x} \right)} }} = \dfrac{1}{{\sqrt {1 - {{\left( x \right)}^2}} }}\]. Hence, the derivative of \[{\sin ^{ - 1}}x\] is given by \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\]. Similarly, we can use this method to find the derivative of other trigonometric functions.