Solveeit Logo
Login

Question

Question: How do you differentiate \(f\left( x \right)=\sin \left( \cos \left( \tan x \right) \right)\) using ...

How do you differentiate f(x)=sin⁡(cos⁡(tan⁡x))f\left( x \right)=\sin \left( \cos \left( \tan x \right) \right) using the chain rule?

Explanation

Solution

In this question we have the given expression in the form of a composite function therefore we will use the chain rule on the terms which is given by f′(x)=g′(h(x))h′(x)f'(x)=g'(h(x))h'(x) which is for two terms. We will modify this formula for three terms accordingly. In this question we have the composite function of three terms and we will use the chain rule from the outer function towards the inner function. In this question we will consider g(x)=sin⁡xg\left( x \right)=\sin x, h(x)=cos⁡xh\left( x \right)=\cos x and i(x)=tan⁡xi\left( x \right)=\tan x, and use the chain rule and simplify the terms to get the required solution.

Complete step-by-step solution:
We have the given equation as:
⇒f(x)=sin⁡(cos⁡(tan⁡x))\Rightarrow f\left( x \right)=\sin \left( \cos \left( \tan x \right) \right)
Since we have to find the derivative of the term, it can be written as:
⇒f′(x)=ddxsin⁡(cos⁡(tan⁡x))\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\sin \left( \cos \left( \tan x \right) \right)
Now since there is no direct formula for calculating the derivative of the given expression, we will use the chain rule.
Consider the outer function g(x)=sin⁡xg\left( x \right)=\sin x, we know that ddxsin⁡x=cos⁡x\dfrac{d}{dx}\sin x=\cos x therefore on using the formula and applying chain rule, we get:
⇒f′(x)=cos⁡(cos⁡(tan⁡x))ddxcos⁡(tan⁡x)\Rightarrow f'\left( x \right)=\cos \left( \cos \left( \tan x \right) \right)\dfrac{d}{dx}\cos \left( \tan x \right)
Consider the next outer function h(x)=cos⁡xh\left( x \right)=\cos x, we know that ddxcos⁡x=−sin⁡x\dfrac{d}{dx}\cos x=-\sin x therefore on using the formula and applying chain rule, we get:
⇒f′(x)=cos⁡(cos⁡(tan⁡x))×−sin⁡(tan⁡x)ddxtan⁡x\Rightarrow f'\left( x \right)=\cos \left( \cos \left( \tan x \right) \right)\times -\sin \left( \tan x \right)\dfrac{d}{dx}\tan x
On simplifying, we get:
⇒f′(x)=−cos⁡(cos⁡(tan⁡x))sin⁡(tan⁡x)ddxtan⁡x\Rightarrow f'\left( x \right)=-\cos \left( \cos \left( \tan x \right) \right)\sin \left( \tan x \right)\dfrac{d}{dx}\tan x
Now consider the innermost function i(x)=tan⁡xi\left( x \right)=\tan x, we know that ddxtan⁡x=sec⁡2x\dfrac{d}{dx}\tan x={{\sec }^{2}}x therefore on using the formula, we get:
⇒f′(x)=−cos⁡(cos⁡(tan⁡x))sin⁡(tan⁡x)sec⁡2x\Rightarrow f'\left( x \right)=-\cos \left( \cos \left( \tan x \right) \right)\sin \left( \tan x \right){{\sec }^{2}}x, which is the required solution.

Note: The various differentiation formulas should be remembered for doing these types of questions. It is to be remembered that the property of chain rule is only unique to differentiation and not integration and therefore the chain rule should not be used in integration questions. Also, the product rule is different for integration questions.