Question

please give answer please don't spam I will report who give right answer I mark brainlist​

Answer 1

Answer:

Step-by-step explanation:

use quotient rule

2cosx(-sin2x)-cos2x(-sinx) / (cos^2x)

=(-2cosxsin2x+sinxcos2x) / cos^2x

Answer 2

$\large\underline{ \sf{Given \:Question - }}$

$\rm :\longmapsto\:Find \: \dfrac{d}{dx}\bigg[\dfrac{cos2x}{cosx} \bigg] \:$

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:y = \dfrac{cos2x}{cosx}$

We know,

$\boxed{ \bf{ \: cos2x = 2 {cos}^{2}x - 1}}$

So, using this, we get

$\rm :\longmapsto\:y = \dfrac{ {2cos}^{2}x - 1}{cosx}$

$\rm :\longmapsto\:y = \dfrac{ {2cos}^{2}x}{cosx} - \dfrac{1}{cosx}$

$\rm :\longmapsto\:y = 2cosx - secx$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( 2cosx - secx)$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx}2cosx - \dfrac{d}{dx}secx$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)}}$

and

$\boxed{ \bf{ \: \dfrac{d}{dx}secx = secx \: tanx}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 2\dfrac{d}{dx}cosx \: - \: secx \: tanx$

$\bf :\longmapsto\:\dfrac{dy}{dx} = - 2 \: sinx \: - \: secx \: tanx$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$