Find y' for y = 1/(3 cos³x) - 1/(cos x)
Find $y'$ for $y = \dfrac{1}{3\cos^{3}x} - \dfrac{1}{\cos x}$.
Explanation of Method
Two negative powers of cosine whose derivatives combine into one term.
Step 1: Rewrite as powers
\[ y = \tfrac13\bigl(\cos x\bigr)^{-3} - \bigl(\cos x\bigr)^{-1}. \]
Step 2: Differentiate by the chain rule
In each term the inner function is $\cos x$, with $(\cos x)' = -\sin x$: \[\frac{d}{dx}\!\left[\tfrac13\bigl(\cos x\bigr)^{-3}\right] = \tfrac13\cdot(-3)\bigl(\cos x\bigr)^{-4}\cdot(-\sin x) = \frac{\sin x}{\cos^{4}x},\]
\[ \frac{d}{dx}\!\left[-\bigl(\cos x\bigr)^{-1}\right] = -(-1)\bigl(\cos x\bigr)^{-2}\cdot(-\sin x) = -\frac{\sin x}{\cos^{2}x}. \]
Step 3: Combine and simplify
Factor out $\dfrac{\sin x}{\cos^{2}x}$ and use $1-\cos^{2}x = \sin^{2}x$: \[y' = \frac{\sin x}{\cos^{4}x} - \frac{\sin x}{\cos^{2}x} = \frac{\sin x}{\cos^{2}x}\!\left(\frac{1}{\cos^{2}x}-1\right) = \frac{\sin x}{\cos^{2}x}\cdot\frac{1-\cos^{2}x}{\cos^{2}x} = \frac{\sin^{3}x}{\cos^{4}x}.\]
Step 4: Verification
\[x = 0:\quad y'(0) = 0, \text{ and } y(0) = \tfrac13 - 1 = -\tfrac23 \text{ is a local maximum} \;\;\checkmark\]
\[x = \frac{\pi}{4}:\quad y' = \frac{(\sqrt2/2)^{3}}{(\sqrt2/2)^{4}} = \frac{1}{\sqrt2/2} = \sqrt{2} \approx 1.41421\]
\[\frac{y(\pi/4+10^{-5})-y(\pi/4-10^{-5})}{2\times 10^{-5}} \approx 1.41421 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{\sin^{3}x}{\cos^{4}x} = \tan^{3}x\,\sec x, \qquad x \neq \frac{\pi}{2}+k\pi\,}\]
Remark
Remark. Read backwards, this is the standard antiderivative $\displaystyle\int\tan^{3}x\sec x\,dx = \tfrac13\sec^{3}x - \sec x + C$. The sign of $y'$ is the sign of $\sin x$, since $\cos^{4}x \gt 0$ wherever the function is defined.