How do you differentiate $\frac{cos x}{1 - sin x}$ $(cos x) / (1-sinx)$ ?

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How do you differentiate $\frac{cos x}{1 - sin x}$ $(cos x) / (1-sinx)$ ?

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Answer 1 · 2016-06-13T15:50:12

Quotient Rule:-

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

$\frac{d y}{d x} = \frac{v \cdot d u - u \cdot d v}{v^{2}}$ $dy/dx=(v*du-u*dv)/v^2$

Let $y = \frac{cos x}{1 - sin x}$ $y=(cosx)/(1-sinx)$

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

$\Rightarrow \frac{d y}{d x} = \frac{(1 - sin x) \frac{d}{d x} (cos x) - cos x \frac{d}{d x} (1 - sin x)}{{(1 - sin x)}^{2}}$ $implies dy/dx=((1-sinx)d/dx(cosx)-cosxd/dx(1-sinx))/(1-sinx)^2$

Since $\frac{d}{d x} (cos x) = - sin x$ $d/dx(cosx)=-sinx$ and $\frac{d}{d x} (1 - sin x) = - cos x$ $d/dx(1-sinx)=-cosx$

Therefore $\frac{d y}{d x} = \frac{(1 - sin x) (- sin x) - cos x (- cos x)}{{(1 - sin x)}^{2}}$ $dy/dx=((1-sinx)(-sinx)-cosx(-cosx))/(1-sinx)^2$

$\Rightarrow \frac{d y}{d x} = \frac{- sin x + {sin}^{2} x + {cos}^{2} x}{{(1 - sin x)}^{2}}$ $implies dy/dx=(-sinx+sin^2x+cos^2x)/(1-sinx)^2$

Since ${sin}^{2} x + {cos}^{2} x = 1$ $Sin^2x+Cos^2x=1$

Therefore $\frac{d y}{d x} = \frac{1 - sin x}{{(1 - sin x)}^{2}} = \frac{1}{1 - sin x}$ $dy/dx=(1-sinx)/(1-sinx)^2=1/(1-Sinx)$

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

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