Find y' for y = x sin x arctan x
Find $y'$ for $y = x\sin x\arctan x$.
Explanation of Method
The three-factor product rule with mixed function types.
Step 1: State the product rule for three factors
Applying the two-factor rule twice gives, for $y = uvw$, $y' = u'vw + uv'w + uvw'$.
Each term differentiates exactly one factor and leaves the other two alone.
Step 2: Differentiate each factor
\[ u = x,\quad u' = 1; \qquad v = \sin x,\quad v' = \cos x; \qquad w = \arctan x,\quad w' = \frac{1}{1+x^{2}}. \]
Step 3: Substitute
\[ y' = 1\cdot\sin x\cdot\arctan x + x\cdot\cos x\cdot\arctan x + x\cdot\sin x\cdot\frac{1}{1+x^{2}}. \]
Step 4: Verification
\[x = 0:\quad y'(0) = 0 + 0 + 0 = 0, \text{ and near the origin } y \approx x^{3} \;\;\checkmark\]
\[x = 1:\quad y' = \sin 1\cdot\frac{\pi}{4} + \cos 1\cdot\frac{\pi}{4} + \frac{\sin 1}{2} \approx 1.5060\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.5060 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \sin x\arctan x + x\cos x\arctan x + \dfrac{x\sin x}{1+x^{2}}\,}\]
Remark
Remark. The first two terms can be grouped as $\bigl(\sin x + x\cos x\bigr)\arctan x$, which is $(x\sin x)'\arctan x$ — the three-factor rule seen as the two-factor rule applied to $(x\sin x)\cdot\arctan x$. Both readings give the same answer.