$\text { Use the Product Rule to differentiate } f ( u ) = \sqrt { u } \left( 5 - u ^ { 6 } \right) \text {. }$
A) $f ^ { t } ( u ) = 6 u ^ { \frac { 11 } { 2 } } + \frac { 5 - u ^ { 7 } } { 2 \sqrt { u } }$
B) $f ^ { t } ( u ) = - 6 u ^ { \frac { 13 } { 2 } } - \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$
C) $f ^ { t } ( u ) = 6 u ^ { \frac { 13 } { 2 } } - \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$
D) $f ^ { t } ( u ) = - 6 u ^ { \frac { 11 } { 2 } } + \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$
E) $f ^ { t } ( u ) = - 6 u ^ { \frac { 13 } { 2 } } + \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$

Question

$\text { Use the Product Rule to differentiate } f ( u ) = \sqrt { u } \left( 5 - u ^ { 6 } \right) \text {. }$ 
A) $f ^ { t } ( u ) = 6 u ^ { \frac { 11 } { 2 } } + \frac { 5 - u ^ { 7 } } { 2 \sqrt { u } }$
B) $f ^ { t } ( u ) = - 6 u ^ { \frac { 13 } { 2 } } - \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$
C) $f ^ { t } ( u ) = 6 u ^ { \frac { 13 } { 2 } } - \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$
D) $f ^ { t } ( u ) = - 6 u ^ { \frac { 11 } { 2 } } + \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$
E) $f ^ { t } ( u ) = - 6 u ^ { \frac { 13 } { 2 } } + \frac { 5 - u ^ { 6 } } { 2 \sqrt { u } }$

Quizplus · Accepted Answer

Using the product rule, we have:

$$f'(u) = \frac{d}{du} (\sqrt{u}) \cdot (5-u^6) + \sqrt{u} \cdot \frac{d}{du}(5-u^6)$$

$$= \frac{1}{2\sqrt{u}} \cdot (5-u^6) + \sqrt{u} \cdot (-6u^5)$$

$$= -3u^{\frac{11}{2}} + \frac{5-u^6}{2u^{\frac{1}{2}}}$$

Simplifying this, we get:

$$f'(u) = -6u^{\frac{11}{2}} + \frac{5-u^6}{2u^{\frac{1}{2}}}$$

Therefore, the answer is D.

Use the Product Rule to differentiate f(u)=u(5−u6). \text { Use the Product Rule to differentiate } f ( u ) = \sqrt { u } \left( 5 - u ^ { 6 } \right) \text {. } Use the Product Rule to differentiate f(u)=u​(5−u6).

$\text { Use the Product Rule to differentiate } f ( u ) = \sqrt { u } \left( 5 - u ^ { 6 } \right) \text {. }$