Question

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.

Answer 1

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