# How do you differentiate (cos x) / (1-sinx)?

Jun 13, 2016

Quotient Rule:-

If $u$ and $v$ are two differentiable functions at $x$ with $v \ne 0$, then $y = \frac{u}{v}$ is differentiable at $x$ and

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \cdot \mathrm{du} - u \cdot \mathrm{dv}}{v} ^ 2$

Let $y = \frac{\cos x}{1 - \sin x}$

Differentiate w.r.t. 'x' using quotient rule

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \sin x\right) \frac{d}{\mathrm{dx}} \left(\cos x\right) - \cos x \frac{d}{\mathrm{dx}} \left(1 - \sin x\right)}{1 - \sin x} ^ 2$

Since $\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$ and $\frac{d}{\mathrm{dx}} \left(1 - \sin x\right) = - \cos x$

Therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \sin x\right) \left(- \sin x\right) - \cos x \left(- \cos x\right)}{1 - \sin x} ^ 2$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin x + {\sin}^{2} x + {\cos}^{2} x}{1 - \sin x} ^ 2$

Since $S {\in}^{2} x + C o {s}^{2} x = 1$

Therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \sin x}{1 - \sin x} ^ 2 = \frac{1}{1 - S \in x}$

Hence, derivative of the given expression is $\frac{1}{1 - \sin x} .$