Question

find $\frac{dy}{dx}$ for $y = \frac{sin x}{cos^{2}x}$. $\frac{dy}{dx}=square$

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, $u = \sin x$, $u'=\cos x$, $v=\cos^{2}x$, and $v'=- 2\cos x\sin x$.

Step2: Substitute values into quotient - rule

$\frac{dy}{dx}=\frac{\cos x\cdot\cos^{2}x-\sin x\cdot(-2\cos x\sin x)}{\cos^{4}x}$.

Step3: Simplify the numerator

$\cos x\cdot\cos^{2}x-\sin x\cdot(-2\cos x\sin x)=\cos^{3}x + 2\sin^{2}x\cos x$.

Step4: Factor out $\cos x$ from the numerator

$\cos^{3}x + 2\sin^{2}x\cos x=\cos x(\cos^{2}x + 2\sin^{2}x)$.

Step5: Use the identity $\sin^{2}x+\cos^{2}x = 1$

$\frac{\cos x(\cos^{2}x + 2\sin^{2}x)}{\cos^{4}x}=\frac{\cos^{2}x+2\sin^{2}x}{\cos^{3}x}=\frac{\cos^{2}x+\sin^{2}x+\sin^{2}x}{\cos^{3}x}=\frac{1 + \sin^{2}x}{\cos^{3}x}$.

Answer:

$\frac{1+\sin^{2}x}{\cos^{3}x}$