Question

Find $\frac{dy}{dx}$, if $y={\left(\sin x\right)}^{\tan x}+{\left(\cos x\right)}^{\sec x}.$

• A. $\left(\sin x{)}^{\tan x}\right(1+{\sec }^{2}x\ln \left(\sin x\right))+(\cos x{)}^{\sec x}\left(\sec x\tan x\right(\ln \cos x-1\left)\right)$
• B. $\left(\sin x{)}^{\tan x}\right(1-{\sec }^{2}x\ln \left(\sin x\right))+(\cos x{)}^{\sec x}\left(\sec x\tan x\right(\ln \cos x-1\left)\right)$
• C. $\left(\sin x{)}^{\tan x}\right(1-{\sec }^{2}x\ln \left(\sin x\right))-(\cos x{)}^{\sec x}\left(\sec x\tan x\right(\ln \cos x-1\left)\right)$
• D. $\left(\sin x{)}^{\tan x}\right(1-{\sec }^{2}x\ln \left(\sin x\right))-(\cos x{)}^{\sec x}\left(\sec x\tan x\right(\ln \cos x+1\left)\right)$

Answer 1

The correct option is A $\left(\sin x{)}^{\tan x}\right(1+{\sec }^{2}x\ln \left(\sin x\right))+(\cos x{)}^{\sec x}\left(\sec x\tan x\right(\ln \cos x-1\left)\right)$

$y=(\sin x{)}^{\tan x}+(\cos x{)}^{\sec x}$

$y=u+v$

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$u=(\sin x{)}^{\tan x}$

$\ln u=\tan x\ln \sin x$

$\frac{1}{4}\frac{du}{dx}=\frac{\tan x\times \cos x}{\sin x}+{\sec }^{2}x\ln \sin x$

$\frac{du}{dx}=\left(\sin x{)}^{\tan x}\right(1+{\sec }^{2}x\ln \sin x)$

$v=(\cos x{)}^{\sec x}\ln v=\sec x\ln \cos x$

$\frac{1}{v}\frac{dv}{dx}=\sec x\tan x\ln \cos x+\frac{\sec x}{\cos x}(-\sin x)$

$=\sec x\tan x(\ln \cos x-1)$

$\frac{dv}{dx}=\left(\cos x{)}^{\sec x}\right(\sec x\tan x(\ln \cos x-1))$

$\frac{dy}{dx}=\left(\sin x{)}^{\tan x}\right(1+{\sec }^{2}x\ln \sin x)+(\cos x{)}^{\sec x}\left(\sec x\tan x\right)$

$(\ln \cos x-1))$