Question

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

Answer 1

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.