Find y .
-$$y = \left( 7 + \frac { 3 } { x } \right) ^ { 4 }$$
A) $- \frac { 12 } { x ^ { 2 } } \left( 7 + \frac { 3 } { x } \right) ^ { 3 }$
B) $\frac { 108 } { x ^ { 4 } } \left( 7 + \frac { 3 } { x } \right) ^ { 2 } + \frac { 24 } { x ^ { 3 } } \left( 7 + \frac { 3 } { x } \right) ^ { 3 }$
C) $12 \left( 7 + \frac { 3 } { x } \right) ^ { 2 }$
D) $- \frac { 36 } { x ^ { 2 } } \left( 7 + \frac { 3 } { x } \right) ^ { 2 } + \frac { 24 } { x ^ { 3 } } \left( 7 + \frac { 3 } { x } \right) ^ { 3 }$

Question

Quizplus · Accepted Answer

Using the chain rule, we have:$$\frac{dy}{dx} = 4(7 + \frac{3}{x})^3 \left(\frac{d}{dx}(7 + \frac{3}{x})\right)$$$$= 4(7 + \frac{3}{x})^3 \left(-\frac{3}{x^2}\right)$$$$= \frac{108}{x^4}(7 + \frac{3}{x})^3 + \frac{24}{x^3}(7 + \frac{3}{x})^3$$

Find Y - y=(7+3x)4y = \left( 7 + \frac { 3 } { x } \right) ^ { 4 }y=(7+x3​)4

Find Y
- $y = \left( 7 + \frac { 3 } { x } \right) ^ { 4 }$