Question

How do you differentiate $f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right)$ ?

Answer 1

Hint: To solve this question, we use the chain rule to differentiate. As we know that the differentiation of inverse tangent is given as $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$. We substitute the values and also multiply it with the differentiation of the function within.

Formula used: $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$

Complete step by step solution:
In this question, we are asked to differentiate an inverse trigonometric function given as $f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right)$.
We follow the chain rule to solve this as there is another function inside the inverse trigonometric function.
The chain rule goes as follows:$f\left( x \right) = g\left( {h\left( x \right)} \right)$
$ \Rightarrow f\left( x \right) = g'\left( {h\left( x \right)} \right) \times h'\left( x \right)$
Thus differentiating our function with respect to $x$, we get:
\[ \Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{1}{{1 + {{\left( {\dfrac{x}{2}} \right)}^2}}} \times \dfrac{{d\left( {\dfrac{1}{2}x} \right)}}{{dx}}\]
As the differentiation of inverse tangent function is given as $\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$
\[ \Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{1}{{1 + \dfrac{{{x^2}}}{4}}} \times \dfrac{1}{2}\]
Simplifying it further, we get:
\[ \Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{{x^2}}}{4}}}\]
Let us take the LCM of the denominator:
\[ \Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{4 + {x^2}}}{4}}}\]
On simplifying it further, we get:
\[ \Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{1}{2} \times \dfrac{4}{{4 + {x^2}}}\]
On cancel the term and we get,
\[ \Rightarrow \dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{1}{{\not{2}}} \times \dfrac{{\not{4}}}{{4 + {x^2}}}\]
Let us rewriting the term and we get

Thus, \[\dfrac{{d\left( {{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}{{dx}} = \dfrac{2}{{4 + {x^2}}}\]

Note: Derivatives are a fundamental tool of Calculus, which is a branch of Mathematics. Derivation or differentiation is a way of finding the instantaneous rate of change of a function based on one of its variables. We can find the derivative of trigonometric functions, logarithmic functions, etc using different formulas. The inverse of differentiation is integration. While differentiation means to break down a function into parts, integration means to add up the infinitesimal parts to give a whole function.
Some common formulas of differentiation are:
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) + g\left( x \right)\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$
 The above formula is for the product rule of differentiation.
Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot.