Find $y ^ { \prime }$ . $y = \left[ \ln \left( x ^ { 9 } + 5 \right) \right] ^ { 3 }$
A) $\frac { 27 x ^ { 8 } \ln \left( x ^ { 9 } + 5 \right) ^ { 2 } } { \left( x ^ { 9 } + 5 \right) }$
B) $\frac { 3 x ^ { 7 } \ln \left( x ^ { 9 } + 5 \right) ^ { 2 } } { \left( x ^ { 9 } + 5 \right) }$
C) $- \frac { 3 x ^ { 8 } \ln \left( x ^ { 9 } + 5 \right) ^ { 2 } } { \left( x ^ { 9 } + 5 \right) }$
D) $- \frac { 9 x ^ { 8 } \ln \left( x ^ { 9 } + 5 \right) ^ { 2 } } { \left( x ^ { 9 } + 5 \right) }$
E) $- \frac { 54 x ^ { 7 } \ln \left( x ^ { 9 } + 5 \right) ^ { 2 } } { \left( x ^ { 9 } + 5 \right) }$

Question

Quizplus · Accepted Answer

To find $y'$, we use the chain rule and the power rule. First, differentiate the outer function $\left[ \ln(x^9 + 5) \right]^3$ with respect to $\ln(x^9 + 5)$, which gives $3\left[ \ln(x^9 + 5) \right]^2$. Then, differentiate $\ln(x^9 + 5)$ with respect to $x^9 + 5$, which gives $\frac{1}{x^9 + 5}$. Finally, differentiate $x^9 + 5$ with respect to $x$, which gives $9x^8$. Multiply these derivatives together: $3\left[ \ln(x^9 + 5) \right]^2 \cdot \frac{1}{x^9 + 5} \cdot 9x^8 = \frac{27x^8 \ln(x^9 + 5)^2}{x^9 + 5}$.

Find y′y ^ { \prime }y′ y=[ln⁡(x9+5)]3y = \left[ \ln \left( x ^ { 9 } + 5 \right) \right] ^ { 3 }y=[ln(x9+5)]3

Find $y ^ { \prime }$ $y = \left[ \ln \left( x ^ { 9 } + 5 \right) \right] ^ { 3 }$