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

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