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

Question 99

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Find the indicated derivative. The independent variable is a function of t. y=x0.7(1+x) ;dy dt=y = x ^ { 0.7 } ( 1 + x ) ; \frac { \mathrm { d } y } { \mathrm {~d} t } =


A) dy dt=(0.7x0.3) dx dt\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 0.7 x ^ { - 0.3 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }
B) dy dt=(0.7x0.3+2.7x0.7) dx dt\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) dy dt=(0.7x0.3+1.7x0.7) dx dt\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) dy dt=(1.7x0.7) dx dt\frac { \mathrm { d } y } { \mathrm {~d} t } = \left( 1.7 x ^ { 0.7 } \right) \frac { \mathrm { d } x } { \mathrm {~d} t }
E) dy dt=(0.7x0.7+2.7x0.7) dx dt\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 }

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