Find the indicated derivative. The independent variable is a function of t. $y = x ^ { 0.7 } ( 1 + x ) ; \frac { \mathrm { d } y } { \mathrm {~d} t } = $
A) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
B) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } + 2.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
C) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } + 1.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
D) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 1.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
E) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { 0.7 } + 2.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$

Question

Find the indicated derivative. The independent variable is a function of t.   $y = x ^ { 0.7 } ( 1 + x ) ; \frac { \mathrm { d } y } { \mathrm {~d} t } =  $  
A) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
B) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } + 2.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
C) $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } + 1.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
D)   $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 1.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$
E)  $\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { 0.7 } + 2.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }$

Quizplus · Accepted Answer

Using the product rule and chain rule, the derivative of $y = x^{0.7}(1 + x)$ with respect to $t$ is $\frac{dy}{dt} = 0.7x^{-0.3}(1 + x) + x^{0.7}(1) = (0.7x^{-0.3} + 1.7x^{0.7})\frac{dx}{dt}$.

Find the Indicated Derivative y=x0.7(1+x);dy dt=y = x ^ { 0.7 } ( 1 + x ) ; \frac { \mathrm { d } y } { \mathrm {~d} t } = y=x0.7(1+x); dtdy​=

Find the Indicated Derivative $y = x ^ { 0.7 } ( 1 + x ) ; \frac { \mathrm { d } y } { \mathrm {~d} t } =$