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Differentiate in Two Ways, Using Product Rule and Otherwise, the Function (1 + 2 Tan X) (5 + 4 Cos X). Verify that the Answers Are the Same. - Mathematics

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Differentiate in two ways, using product rule and otherwise, the function (1 + 2 tan x) (5 + 4 cos x). Verify that the answers are the same. 

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Solution

\[{\text{ Product rule } (1}^{st} \text{ method }):\]
\[\text { Let } u = 1 + 2 \tan x; v = 5 + 4 \cos x\]
\[\text{ Then }, u' = 2 \sec^2 x; v' = - 4 \sin x\]
\[\text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uv \right) = uv' + vu'\]
\[\frac{d}{dx}\left[ \left( 1 + 2 \tan x \right)\left( 5 + 4 \cos x \right) \right] = \left( 1 + 2 \tan x \right)\left( - 4 \sin x \right) + \left( 5 + 4 \cos x \right)\left( 2 \sec^2 x \right)\]
\[ = - 4 \sin x - 8 \tan x \sin x + 10 \sec^2 x + 8 \sec x\]
\[ = - 4 \sin x + 10 \sec^2 x + \left( \frac{8}{\cos x} - \frac{8 \sin^2 x}{\cos x} \right)\]
\[ = - 4 \sin x + 10 \sec^2 x + 8\left( \frac{1 - \sin^2 x}{\cos x} \right)\]
\[ = - 4 \sin x + 10 \sec^2 x + 8\left( \frac{\cos^2 x}{\cos x} \right)\]
\[ = - 4 \sin x + 10 \sec^2 x + 8 \cos x\]
\[ 2^{nd} \text{ method }:\]
\[\left( 1 + 2 \tan x \right)\left( 5 + 4 \cos x \right) = 5 + 4 \cos x + 10 \tan x + 8 \sin x\]
\[\text{ Now, we have }:\]
\[\frac{d}{dx}\left[ \left( 1 + 2 \tan x \right)\left( 5 + 4 \cos x \right) \right] = \frac{d}{dx}\left( 5 + 4 \cos x + 10 \tan x + 8 \sin x \right)\]
\[ = - 4 \sin x + 10 \sec^2 x + 8 \cos x\]
\[\text{ Using both the methods, we get the same answer } .\]

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.4 | Q 25 | Page 39

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