Question

What is the derivative of $\arctan \left( {\dfrac{1}{x}} \right)$?

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
$y = \arctan \left( {\dfrac{1}{x}} \right)$
Or
$y = ta{n^{ - 1}}\left( {\dfrac{1}{x}} \right) - - - (1)$.
We know that the differentiation of ${\tan ^{ - 1}}x$ with respect to ‘x’ is $\dfrac{1}{{1 + {x^2}}}$. That is
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$.
Or
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}\dfrac{{dx}}{{dx}} - - - (2)$. Both are the same.
Now let’s put $u = \dfrac{1}{x}$ in equation 1
$\Rightarrow y = ta{n^{ - 1}}\left( u \right)$
Differentiate this with respect to ‘x’,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}ta{n^{ - 1}}\left( u \right)$
Using equation (2), we will get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {u^2}}}\dfrac{{du}}{{dx}}$
But we have taken $u = \dfrac{1}{x}$, then above becomes
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {\dfrac{1}{x}} \right)}^2}}}\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$
We know that the differentiation of $\dfrac{1}{x}$ is $- \dfrac{1}{{{x^2}}}$. That is $\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}$.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {\dfrac{1}{x}} \right)}^2}}} \times \dfrac{{ - 1}}{{{x^2}}}$
Again simplifying we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + \dfrac{1}{{{x^2}}}}} \times \dfrac{{ - 1}}{{{x^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\dfrac{{1 + {x^2}}}{{{x^2}}}}} \times \dfrac{{ - 1}}{{{x^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^2}}}{{1 + {x^2}}} \times \dfrac{{ - 1}}{{{x^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{1 + {x^2}}}$. This is the required answer.
That is differentiation of $\arctan \left( {\dfrac{1}{x}} \right)$ with respect to ‘x’ 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.