Find y' for y = tan x
Find $y'$ for $y = \tan x$.
Explanation of Method
Derive the derivative of the tangent from the quotient rule.
Step 1: Rewrite as a quotient
\[ y = \tan x = \frac{\sin x}{\cos x}. \]
Step 2: Apply the quotient rule
For $y = \dfrac{u}{v}$ the quotient rule gives $y' = \dfrac{u'v - uv'}{v^{2}}$. Here $u = \sin x$ and $v = \cos x$, so $u' = \cos x$ and $v' = -\sin x$: \[y' = \frac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^{2} x} = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}.\]
Step 3: Apply the Pythagorean identity
Using $\sin^{2} x + \cos^{2} x = 1$: $y' = \frac{1}{\cos^{2} x}$.
Step 4: Verification
\[\frac{1}{\cos^{2} x} = \sec^{2} x, \text{ the standard table form} \;\;\checkmark\]
\[x = 0:\quad y'(0) = \frac{1}{\cos^{2} 0} = 1, \text{ matching the tangent line } y = x \text{ to } \tan x \text{ at the origin} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \frac{1}{\cos^{2} x}, \qquad x \neq \frac{\pi}{2}+k\pi,\; k \in \mathbb{Z}\,}\]
Remark
Remark. Since $\dfrac{1}{\cos^{2}x} \geq 1 \gt 0$ everywhere it is defined, $\tan x$ is strictly increasing on each interval between consecutive vertical asymptotes. An equivalent form is $y' = 1 + \tan^{2} x$, which follows from dividing the Pythagorean identity by $\cos^{2} x$.