Q. The derivative of $a^{\sec x}w.r.t.a^{\tan x}(a > 0)$ is
A) ${\sec{x\ a}}^{\sec{x - \tan x}}$
B) ${\sin{x\ a}}^{\tan{x - \sec x}}$
C) ${\sin{x\ a}}^{\sec{x - \tan x}}$
D) $a^{\sec{x - \tan x}}$
Explanation
(c)
Let $u = a^{\sec x}\ and\ v = a^{\tan x}$
$\Rightarrow \ \ \ \frac{du}{dx} = a^{\sec x}\log_{e}{a.\sec{x\tan x}}\ \$
and $\ \frac{dv}{dx} = \ a^{\tan x}\log_{e}{a.\sec^{2}x}$
$\therefore\ \frac{du}{dv} = \frac{du\ /dx}{dv\ /dx} = \frac{a^{\sec x}\log_{e}a\sec{x\tan x}}{a^{\tan x}\log_{e}a\sec^{2}x}\$
$= \ a^{\sec{x - \tan x}}\sin x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.