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Question: Differentiate \(y=\dfrac{\tan x}{x}\)?...

Differentiate y=tan⁡xxy=\dfrac{\tan x}{x}?

Explanation

Solution

We first define the division rule and how the differentiation of function works. We take addition of these two different differentiated values. We take the dydx\dfrac{dy}{dx} altogether. We keep one function and differentiate the other one and then do the same thing with the other function. Then we take the addition to complete the formula.

Complete step by step answer:
We now discuss the multiplication process of two functions where f(x)=u(x)v(x)f\left( x \right)=\dfrac{u\left( x \right)}{v\left( x \right)}
Differentiating f(x)=u(x)v(x)f\left( x \right)=\dfrac{u\left( x \right)}{v\left( x \right)}, we get ddx[f(x)]=ddx[uv]=vdudx−udvdxv2\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ \dfrac{u}{v} \right]=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}.
The above-mentioned rule is the division rule. We apply that on y=tan⁡xxy=\dfrac{\tan x}{x}. We assume the functions where u(x)=tan⁡x,v(x)=xu\left( x \right)=\tan x,v\left( x \right)=x

We know that differentiation of u(x)=tan⁡xu\left( x \right)=\tan x is u′(x)=sec⁡2x{{u}^{'}}\left( x \right)={{\sec }^{2}}x and differentiation of v(x)=xv\left( x \right)=x is v′(x)=1{{v}^{'}}\left( x \right)=1. We now take differentiation on both parts of y=tan⁡xxy=\dfrac{\tan x}{x} and get dydx=ddx[tan⁡xx]\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \dfrac{\tan x}{x} \right].
We place the values of u′(x)=sec⁡2x{{u}^{'}}\left( x \right)={{\sec }^{2}}x and v′(x)=1{{v}^{'}}\left( x \right)=1 to get ddx[tan⁡xx]=xsec⁡2x−tan⁡x×1x2\dfrac{d}{dx}\left[ \dfrac{\tan x}{x} \right]=\dfrac{x{{\sec }^{2}}x-\tan x\times 1}{{{x}^{2}}}.
We take all the dydx\dfrac{dy}{dx} forms altogether to get

\therefore \dfrac{dy}{dx}=\dfrac{x{{\sec }^{2}}x-\tan x}{{{x}^{2}}} $$ **Therefore, differentiation of $y=\dfrac{\tan x}{x}$ is $$\dfrac{dy}{dx}=\dfrac{x{{\sec }^{2}}x-\tan x}{{{x}^{2}}}$$.** **Note:** We need to remember that in the chain rule, we aren’t cancelling out differentials. Canceling the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate. The rule may be extended or generalized to many other situations, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts.