Find $\frac { d y } { d x } \text { if } y = \ln \left( x ^ { 6 } \left( x ^ { 2 } - x + 4 \right) \right)$ .
A) $\frac { 8 x ^ { 2 } - 7 x + 24 } { x \left( x ^ { 2 } - x + 4 \right) }$
B) $\frac { 8 x ^ { 2 } + 7 x - 24 } { x \left( x ^ { 2 } - x + 4 \right) }$
C) $\frac { 8 x ^ { 2 } + 7 x + 6 } { x \left( x ^ { 2 } - x + 4 \right) }$
D) $\frac { 2 x ^ { 2 } - 7 x - 6 } { x \left( x ^ { 2 } - x + 4 \right) }$
E) $\frac { 2 x ^ { 2 } + 7 x + 24 } { x \left( x ^ { 2 } - x + 4 \right) }$

Question

Quizplus · Accepted Answer

Using the chain rule and properties of logarithms, differentiate $y = \ln(x^6(x^2 - x + 4))$. First, rewrite $y$ as $y = \ln(x^6) + \ln(x^2 - x + 4)$. Differentiating gives $\frac{dy}{dx} = \frac{6}{x} + \frac{2x - 1}{x^2 - x + 4}$. Simplifying, we get $\frac{dy}{dx} = \frac{8x^2 - 7x + 24}{x(x^2 - x + 4)}$.

Find dydx if y=ln⁡(x6(x2−x+4))\frac { d y } { d x } \text { if } y = \ln \left( x ^ { 6 } \left( x ^ { 2 } - x + 4 \right) \right)dxdy​ if y=ln(x6(x2−x+4))

Find $\frac { d y } { d x } \text { if } y = \ln \left( x ^ { 6 } \left( x ^ { 2 } - x + 4 \right) \right)$