Question

find the derivative of the following function, where a and b are nonzero constants.
y = (a sin x + b cos x)/(a sin x - b cos x)

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u = a\sin x - b\cos x$, $u'=a\cos x + b\sin x$, $v=a\sin x + b\cos x$, and $v'=a\cos x - b\sin x$.

Step2: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{(a\cos x + b\sin x)(a\sin x + b\cos x)-(a\sin x - b\cos x)(a\cos x - b\sin x)}{(a\sin x + b\cos x)^{2}}$.

Step3: Expand the numerator

\[

$$\begin{align*} &(a\cos x + b\sin x)(a\sin x + b\cos x)-(a\sin x - b\cos x)(a\cos x - b\sin x)\\ =&a^{2}\sin x\cos x+ab\cos^{2}x + ab\sin^{2}x + b^{2}\sin x\cos x-(a^{2}\sin x\cos x - ab\sin^{2}x - ab\cos^{2}x + b^{2}\sin x\cos x)\\ =&a^{2}\sin x\cos x+ab(\cos^{2}x+\sin^{2}x)+b^{2}\sin x\cos x - a^{2}\sin x\cos x+ab(\sin^{2}x+\cos^{2}x)-b^{2}\sin x\cos x\\ =&ab(\cos^{2}x+\sin^{2}x)+ab(\sin^{2}x+\cos^{2}x) \end{align*}$$

\]
Since $\sin^{2}x+\cos^{2}x = 1$, the numerator simplifies to $2ab$.

Answer:

$\frac{2ab}{(a\sin x + b\cos x)^{2}}$