Question

derivative y =(1+tan x).(1-sec x)

Answer 1

Answer:

Hope this helps you in your work.

Answer 2

$\large\bf\underline{\red{Question}}:$

• Derivative y =(1+tan x).(1-sec x)  

$\\\large\bf\underline{\red{Solution}}:$

We know that,

$\\\red\bigstar\;\;\boxed{\bf(UV)' = U'V + V'U}$

$\\\longmapsto \dfrac{d}{dx}\left[(1+\tan x).(1-\sec x)\right]$

$\\\dashrightarrow\dfrac{d}{dx}[(1+\tan x)] . (1-\sec x) + \dfrac{d}{dx}[(1-\sec x)].(1+\tan x)$  

$\\\dashrightarrow \sec^2 x.(1-\sec x) -\sec x \tan x (1+\tan x)$

$\\\dashrightarrow  \pink{\sec^2x-\sec^3x-\sec x\tan x-\sec x\tan ^2x}$

Therefore,

• $\dfrac{d}{dx}[(1+\tan x).(1-\sec x)] =\sec^2x-\sec^3x-\sec x\tan x-\sec x\tan ^2x$  

More to know :

• (tan x)' = sec²x
• (sec x)' = tan x sec x
• (sin x)' = cos x
• (cos x)' = -sin x
• Derivative of any constant is zero (0).