Find y' for y = tan x - 1/3 tan³x + 1/5 tan^5x

Find $y'$ for $y = \tan x - \dfrac{1}{3}\tan^{3}x + \dfrac{1}{5}\tan^{5}x$.

Step 1: Differentiate term by term

Write $t = \tan x$, so that $t' = \dfrac{1}{\cos^{2}x} = \sec^{2}x$. Each term is a power of $t$, handled by the chain rule, and the fractional coefficients cancel the exponents:

\[ \begin{aligned} \frac{d}{dx}\bigl(\tan x\bigr) &= \sec^{2}x, \\[4pt] \frac{d}{dx}\!\left(\tfrac13\tan^{3}x\right) &= \tfrac13\cdot 3\tan^{2}x\cdot\sec^{2}x = \tan^{2}x\,\sec^{2}x, \\[4pt] \frac{d}{dx}\!\left(\tfrac15\tan^{5}x\right) &= \tfrac15\cdot 5\tan^{4}x\cdot\sec^{2}x = \tan^{4}x\,\sec^{2}x. \end{aligned} \]

Step 2: Combine

\[ y' = \sec^{2}x\bigl(1 - \tan^{2}x + \tan^{4}x\bigr). \]

Step 3: Simplify the bracket

The bracket is the factor that turns $1+t^{2}$ into $1+t^{6}$ — it is the sum-of-cubes factorisation with $t^{2}$ in place of $t$: \[\bigl(1+t^{2}\bigr)\bigl(1-t^{2}+t^{4}\bigr) = 1+t^{6} \;\Longrightarrow\; 1-\tan^{2}x+\tan^{4}x = \frac{1+\tan^{6}x}{1+\tan^{2}x} = \frac{1+\tan^{6}x}{\sec^{2}x},\]

using the Pythagorean identity $1+\tan^{2}x = \sec^{2}x$.

Step 4: Substitute back

The two factors $\sec^{2}x$ cancel: $y' = \sec^{2}x\cdot\frac{1+\tan^{6}x}{\sec^{2}x} = 1 + \tan^{6}x$.

Step 5: Verification

\[x = 0:\quad \tan 0 = 0 \;\Longrightarrow\; y' = 1; \qquad \sec^{2}0\,(1-0+0) = 1 \;\;\checkmark\]

\[x = \frac{\pi}{4}:\quad \tan\frac{\pi}{4} = 1 \;\Longrightarrow\; y' = 1+1 = 2; \qquad \sec^{2}\frac{\pi}{4}(1-1+1) = 2\cdot 1 = 2 \;\;\checkmark\]

\[\frac{y(\pi/4+10^{-5})-y(\pi/4-10^{-5})}{2\times 10^{-5}} \approx 2.0 \;\;\checkmark\]