Question

Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.

(3 sec x − 4 cosec x) (−2 sin x + 5 cos x)

Answer 1

\[ {\text{ Product rule } (1}^{st} \text{ method }):\]
\[\text{ Let } u = 3 \sec x - 4 \cos ec x; v = - 2 \sin x + 5 \cos x\]
\[\text{ Then }, u' = 3 \sec x \tan x + 4 cos ec x \cot x; v' = - 2 \cos x - 5 \sin x\]
\[\text{ Using the product rule }:\]
\[\frac{d}{dx}\left( uv \right) = uv' + vu'\]
\[\frac{d}{dx}\left[ \left( 3 sec x - 4 \cos ec x \right)\left( - 2 \sin x + 5 \cos x \right) \right] = \left( 3 sec x - 4 \cosec x \right)\left( - 2 \cos x - 5 \sin x \right) + \left( - 2 \sin x + 5 \cos x \right)\left( 3 \sec x \tan x + 4 \cosec x cot x \right)\]
\[ = - 6 + 15 \tan x + 8 \cot x + 20 - 6 \tan^2 x - 8 cot x - 15 \tan x + 20 \cot^2 x\]
\[ = - 6 + 20 - 6\left( \sec^2 x - 1 \right) + 20 \left( {cosec}^2 x - 1 \right)\]
\[ = - 6 + 20 - 6 \sec^2 x + 6 + 20 {cosec}^2 x - 20\]
\[ = - 6 \sec^2 x + 20 \cos e c^2 x\]
\[ 2^{nd} method:\]
\[\frac{d}{dx}\left[ \left( 3 sec x - 4 \cos ec x \right)\left( - 2 \sin x + 5 \cos x \right) \right] = \frac{d}{dx}\left( - 6 \sec x \sin x + 15 \sec x \cos x + 8 \cos ec x \sin x - 20 \cos ec x \cos x \right)\]
\[ = \frac{d}{dx}\left( - 6 \frac{\sin x}{\cos x} + 15\frac{\cos x}{\cos x} + 8 \frac{\sin x}{\sin x} - 20 \frac{\cos x}{\sin x} \right)\]
\[ = \frac{d}{dx}\left( - 6 \tan x + 15 + 8 - 20 \cot x \right)\]
\[ = \frac{d}{dx}\left( - 6\tan x - 20 \cot x + 23 \right)\]
\[ = - 6 \sec^2 x + 20 \cos e c^2 x\]
\[\text{ Using both the methods, we get the same answer }.\]