Differentiate $y = x \arctan ( 3 x )$ .
A) $y ^ { t } = \frac { 3 x } { \left( 1 + ( 3 x ) ^ { 2 } \right) } + \arctan ( 3 x )$
B) $y ^ { \prime } = \frac { 1 } { 3 + x ^ { 2 } } + 3 \arctan ( 3 x )$
C) $y ^ { \prime } = 3 \arctan ( 3 x )$
D) $y ^ { t } = 1 \cdot \arctan ( 3 x )$

Question

Quizplus · Accepted Answer

To differentiate $y = x \arctan(3x)$, we use the product rule: $(uv)' = u'v + uv'$. Let $u = x$ and $v = \arctan(3x)$. Then, $u' = 1$ and $v' = \frac{3}{1+(3x)^2}$ by the chain rule. Thus, $y' = 1 \cdot \arctan(3x) + x \cdot \frac{3}{1+(3x)^2} = \frac{3x}{1+(3x)^2} + \arctan(3x)$.

Differentiate y=xarctan⁡(3x)y = x \arctan ( 3 x )y=xarctan(3x) A) yt=3x(1+(3x)2)+arctan⁡(3x)y ^ { t } = \frac { 3 x } { \left( 1 + ( 3 x ) ^ { 2 } \right) } + \arctan ( 3 x )yt=(1+(3x)2)3x​+arctan(3x)

Differentiate $y = x \arctan ( 3 x )$
A) $y ^ { t } = \frac { 3 x } { \left( 1 + ( 3 x ) ^ { 2 } \right) } + \arctan ( 3 x )$