find $\frac{dy}{dx}$ for the following function.\n$y = \frac{7cos x}{1+sin x}$\n$\frac{dy}{dx}=square$

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, $u = 7\cos x$, so $u'=-7\sin x$, and $v = 1+\sin x$, so $v'=\cos x$.

Step2: Substitute into quotient - rule formula

$\frac{dy}{dx}=\frac{(-7\sin x)(1 + \sin x)-7\cos x(\cos x)}{(1 + \sin x)^{2}}$.

Step3: Expand the numerator

$\frac{dy}{dx}=\frac{-7\sin x-7\sin^{2}x - 7\cos^{2}x}{(1 + \sin x)^{2}}$.

Step4: Use the identity $\sin^{2}x+\cos^{2}x = 1$

$\frac{dy}{dx}=\frac{-7\sin x-7(\sin^{2}x+\cos^{2}x)}{(1 + \sin x)^{2}}=\frac{-7\sin x - 7}{(1 + \sin x)^{2}}$.

Step5: Simplify the expression

Factor out - 7 from the numerator: $\frac{dy}{dx}=\frac{-7(\sin x + 1)}{(1 + \sin x)^{2}}$. Cancel out the common factor $(\sin x + 1)$ (assuming $\sin x\neq - 1$), we get $\frac{dy}{dx}=-\frac{7}{1+\sin x}$.

Answer:

$-\frac{7}{1+\sin x}$