Question

vi) Using the rule for differentiation for
quotient of two functions, prove that
d /dx ( sinx/cosx)=sec^2 x​

Answer 1

Explanation:

$y = \frac{sinx}{cosx} \\ \frac{dy}{dx} = \frac{cosx \frac{d}{dx} sinx - sinx \frac{d}{dx} cosx}{ {cos}^{2} x} \\ \frac{dy}{dx } = \frac{ {cos}^{2} x + {sin}^{2} x}{ {cos}^{2} x} \\ \frac{dy}{dx} = \frac{1}{ {cos}^{2}x } \\ \frac{dy}{dx} = {sec}^{2} x \\ Hence \: proved$

Answer 2

Solution :

Let, y = sinx / cosx ....(i)

We use the differentiation formula

d/dx (u/v) = (v du/dx - u dv/dx)/v²

where u, v are functions of x

Differentiating both sides of (i), we get

dy/dx = d/dx (sinx / cosx)

= {cosx d/dx (sinx) - sinx d/dx (cosx)}/cos²x

= (cosx * cosx + sinx * sinx)/cos²x

= (cos²x + sin²x) * sec²x

= sec²x

Hence, the required derivative is

d/dx (sinx / cosx) = sec²x (proved)