$\frac{a+b \sin x}{c+d \cos x}$
Question.

$\frac{a+b \sin x}{c+d \cos x}$

solution:

Given $y=\frac{a+b \sin x}{c+d \cos x}$

Applying division rule or quotient rule of differentiation that is

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{t}}{\mathrm{y}}\right)=\frac{\mathrm{y} \cdot \mathrm{dt}}{\mathrm{dx}}-\mathrm{t} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow y=\frac{a+b \sin x}{c+d \cos x}$

$\Rightarrow \frac{d y}{d x}=\frac{(c+d \cos x) \frac{d}{d x}(a+b \sin x)-(a+b \sin x) \frac{d}{d x}(c+d \cos x)}{(c+d \cos x)^{2}}$

On differentiating we get

$\Rightarrow \frac{d y}{d x}=\frac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}}$

$=\left[c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x\right] /(c+d \cos x)^{2}$

$=\left[\operatorname{cbcos} x+a d \sin x+b d\left(\cos ^{2} x+\sin ^{2} x\right)\right] /(c+d \cos x)^{2}$

$=[c b \cos x+a d \sin x+b d] /(c+d \cos x)^{2}$
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