Question

If the derivative of $(x+\sec x)(x-\tan x)$ is $A(x+\sec x)+B(x-\tan x),$ where $A$ and $B$ respectively are

• A. ${\sec }^{2}x-1,1+\sec x\tan x$
• B. $1-{\sec }^{2}x,1-\sec x\tan x$
• C. $1-{\sec }^{2}x,1+\sec x\tan x$
• D. ${\sec }^{2}x-1,1-\sec x\tan x$

Answer 1

Answer: (C)

Let $y=(x+\sec x)(x-\tan x)$
Differentiating $y$ w.r.t. $x$, we get
$\frac{dy}{dx}=(x+\sec x)\frac{d}{dx}(x-\tan x)+(x-\tan x)\frac{d}{dx}(x+\sec x)\text{ (by product rule) }$
$=(x+\sec x)\left(1-{\sec }^{2}x\right)+(x-\tan x)(1+\sec x\tan x)$