Question

What is the derivative of \[y = {\cot ^{ - 1}}\left( x \right)\]?

Answer 1

Hint: Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. First consider the function \[y\], then differentiate \[y\] with respect to \[x\] by using a standard differentiation formula of trigonometric ratio and use chain rule for differentiation. And on further simplification we get the required differentiate value.

Complete step by step solution:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Consider the given function
\[y = {\cot ^{ - 1}}\left( x \right)\]
Now taking cotangent function on both side we will have,
\[\cot y = x\]
Now differentiate this with respect to x
\[\dfrac{d}{{dx}}\cot y = \dfrac{{dx}}{{dx}} - - - (1)\]
Since we know the differentiation of cotangent, that is
\[\dfrac{d}{{dx}}\cot x = - {\csc ^2}x\dfrac{{dx}}{{dx}}\].
Then (1) becomes
\[ - {\csc ^2}y\dfrac{{dy}}{{dx}} = 1\]
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{ - {{\csc }^2}y}}\]
But we know the trigonometric identity \[1 + {\cot ^2}x = {\csc ^2}x\], above becomes
\[\dfrac{{dy}}{{dx}} = - \dfrac{1}{{1 + {{\cot }^2}y}}\]
But in the beginning we have taken \[\cot y = x\], then
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{1 + {x^2}}}\]
Thus, the derivative of \[{\cot ^{ - 1}}\left( x \right)\] is \[ - \dfrac{1}{{1 + {x^2}}}\].

Additional information:
\[ \bullet \]Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as \[h'(x) = af'(x) + bg'(x)\]
\[ \bullet \]Quotient rule: The derivative of one function divided by other is found by quotient rule such as\[{\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]'} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right]}^2}}}\].
\[ \bullet \]Product rule: When a derivative of a product of two function is to be found, then we use product rule that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\].
\[ \bullet \]Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is \[fog'({x_0}) = [(f'og)({x_0})]g'({x_0})\].

Note:
We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.