Question

Find the derivative of $\frac{a+b\sin x}{c+d\cos x}$ where $a,b,c,d$ are fixed non-zero constants.

Answer 1

Let $f\left(x\right)=\frac{a+b\sin x}{c+d\cos x}$
Differentiating with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{a+b\sin x}{c+d\cos x}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(c+d\cos x)\frac{d}{dx}(a+b\sin x)-(a+b\sin x)\frac{d}{dx}(c+d\cos x)}{(c+d\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\left[\right(c+d\cos x\left)\right(b\cos x)-(a+b\sin x\left)\right(-d\sin x\left)\right]}{(c+d\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{[bc\cos x+bd{\cos }^{2}x+ad\sin x+bd{\sin }^{2}x]}{(c+d\cos x{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{bc\cos x+ad\sin x+bd({\cos }^{2}x+{\sin }^{2}x)}{(c+d\cos x{)}^2}$
$\therefore f^{\prime }\left(x\right)=\frac{bc\cos x+ad\sin x+db}{(c+d\cos x{)}^2}$
$[\because {\cos }^{2}x+{\sin }^{2}x=1]$