Find y' for y = (sin² x)/(cos x)
Find $y'$ for $y = \dfrac{\sin^{2} x}{\cos x}$.
Explanation of Method
Differentiate a trigonometric quotient and simplify with the Pythagorean identity.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Using the result of Problem 48 for the numerator: \[u = \sin^{2} x,\quad u' = 2\sin x\cos x, \qquad v = \cos x,\quad v' = -\sin x.\]
Step 2: Substitute
Subtracting $uv'$ with $v' = -\sin x$ adds a positive term: \[y' = \frac{2\sin x\cos x\cdot\cos x - \sin^{2}x\cdot(-\sin x)}{\cos^{2}x} = \frac{2\sin x\cos^{2}x + \sin^{3}x}{\cos^{2}x}.\]
Step 3: Factor the numerator
\[ y' = \frac{\sin x\bigl(2\cos^{2}x + \sin^{2}x\bigr)}{\cos^{2}x}. \]
Split the bracket so that the Pythagorean identity applies: $2\cos^{2}x + \sin^{2}x = \cos^{2}x + \bigl(\cos^{2}x + \sin^{2}x\bigr) = \cos^{2}x + 1$.
Step 4: Verification
\[x = \frac{\pi}{6}:\quad \sin x = \frac12,\ \cos^{2}x = \frac34 \;\Longrightarrow\; y' = \frac{\frac12\left(1+\frac34\right)}{\frac34} = \frac{\frac78}{\frac34} = \frac{7}{6}\]
\[\frac{y(\pi/6 + 10^{-5}) - y(\pi/6 - 10^{-5})}{2\times 10^{-5}} \approx 1.16667 = \frac{7}{6} \;\;\checkmark\]
\[x = 0:\quad y'(0) = 0, \text{ matching the minimum } y(0) = 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{\sin x\bigl(1+\cos^{2}x\bigr)}{\cos^{2}x}, \qquad x \neq \frac{\pi}{2}+k\pi,\; k\in\mathbb{Z}\,}\]
Remark
Remark. An equivalent form is $y = \sec x - \cos x$, since $\dfrac{\sin^{2}x}{\cos x} = \dfrac{1-\cos^{2}x}{\cos x}$; differentiating that gives $\sec x\tan x + \sin x$, which is the same expression written differently. The function is undefined exactly where $\cos x = 0$.