Question

What is the derivative of $\dfrac{1}{ab}{{\tan }^{-1}}\left( \left( \dfrac{b}{a} \right)\tan x \right)$ ?

Answer 1

For solving this question, we first need to have a clear idea of what derivative of a function means. The derivative of composite functions needs to performed using chain rule and step by step we get the required derivation $=\dfrac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}$ .

Complete step-by-step solution:
For solving this question, we first need to have a clear idea of what derivative of a function means. In mathematics, derivative of a function at a chosen input value is the slope of the tangent to that function at that point. In Leibniz’s notation, an infinitesimal change in x is denoted by dx, and the derivative of y with respect to x is denoted by,
$\dfrac{dy}{dx}$ , which denoted the ratio of two infinitesimal values. Now, from the predefined formula of the derivatives of various functions we know that,
$\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ , and $\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$.
In case of the derivation of composite functions (for example $h\left( x \right)=f\left( g\left( x \right) \right)$ is a composite function), we need to apply chain rule. Thus, according to the predefined chain rule formula, we can write,
$\dfrac{d}{dx}h\left( x \right)=\dfrac{d}{dx}f\left( g\left( x \right) \right)=\dfrac{d}{d\left( g\left( x \right) \right)}f\left( g\left( x \right) \right)\times \dfrac{d}{dx}g\left( x \right)$ .
Now solving the given problem according to chain rule and the standard derivative of ${{\tan }^{-1}}x$ ,
$\left( \dfrac{1}{ab} \right)\dfrac{d}{dx}{{\tan }^{-1}}\left( \left( \dfrac{b}{a} \right)\tan x \right)=\dfrac{d}{d\left( \left( \dfrac{b}{a} \right)\tan x \right)}\left( {{\tan }^{-1}}\left( \left( \dfrac{b}{a} \right)\tan x \right) \right)\times \dfrac{d}{dx}\left( \left( \dfrac{b}{a} \right)\tan x \right)$
In the above equation, we differentiated the composite function using chain rule. Applying the predefined formula, we get,

& \Rightarrow \left( \dfrac{1}{ab} \right)\dfrac{d}{dx}{{\tan }^{-1}}\left( \left( \dfrac{b}{a} \right)\tan x \right)=\left( \dfrac{1}{ab} \right)\left\\{ \dfrac{1}{1+{{\left( \left( \dfrac{b}{a} \right)\tan x \right)}^{2}}} \right\\}\times \left( \dfrac{b}{a}{{\sec }^{2}}x \right) \\\ & \Rightarrow \left( \dfrac{1}{ab} \right)\dfrac{d}{dx}{{\tan }^{-1}}\left( \left( \dfrac{b}{a} \right)\tan x \right)=\left( \dfrac{1}{ab} \right)\left( \dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}{{\tan }^{2}}x} \right)\times \left( \dfrac{b}{a}{{\sec }^{2}}x \right) \\\ \end{aligned}$$ Now, according to the formulae of trigonometry, as we know $\tan x=\dfrac{\sin x}{\cos x}$ and $\operatorname{secx}=\dfrac{1}{\cos x}$ , so substituting these values in the above equation we get, $$\Rightarrow \left( \dfrac{1}{ab} \right)\dfrac{d}{dx}{{\tan }^{-1}}\left( \left( \dfrac{b}{a} \right)\tan x \right)=\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{{{a}^{2}}+{{b}^{2}}\left( \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)}=\dfrac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}$$ **Thus, the required derivative to the given problem is $$=\dfrac{1}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}$$.** **Note:** We need to be very careful while solving the derivative of these composite functions. The chain rule for solving these problems is very simple but prone to mistakes. A single miscalculation can lead to a totally different answer. The predefined derivative formulae must also be memorized and written correctly while solving them.