Question

y=sinx/cosx find dy /dx​

Answer 1

Given ,

The function is y = sin(x)/cos(x)

Differentiating y wrt x , we get

$: \mapsto \tt \frac{dy}{dx} = \frac{ \cos(x) \frac{d \sin(x) }{dx} - \sin(x) \frac{d \cos(x) }{dx} }{ { \cos}^{2}(x) }$

$: \mapsto \tt \frac{dy}{dx} = \frac{ \cos(x) \cos(x) + \sin(x) \sin(x) }{ { \cos}^{2}(x) }$

$: \mapsto \tt \frac{dy}{dx} = \frac{ { \cos}^{2}(x) + { \sin}^{2}(x) }{ { \cos}^{2}(x) }$

$: \mapsto \tt \frac{dy}{dx} = \frac{1}{ { \cos}^{2}(x) }$

$: \mapsto \tt \frac{dy}{dx} = {sec}^{2} (x)$

The derivative of given function is sec²(x)

Remmember :

$: \mapsto \tt \frac{d(u.v)}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {(v)}^{2} }$

$: \mapsto \tt \frac{d \sin(x) }{dx} = \cos(x)$

$: \mapsto \tt \frac{d \cos(x) }{dx} = - \sin(x)$

$: \mapsto \tt { \cos}^{2}(x) + { \sin}^{2} (x) = 1$