Find $y ^ { \prime }$ . $y = \log _ { 3 } \left( 4 - x - x ^ { 8 } \right)$
A) $\frac { 8 x ^ { 9 } + x } { ( \ln 3 ) \left( x ^ { 8 } + x - 4 \right) }$
B) $\frac { x ^ { 7 } + x } { ( \ln 3 ) \left( x ^ { 8 } + x - 4 \right) }$
C) $\frac { 8 x ^ { 7 } + x } { \left( x ^ { 8 } + x - 4 \right) }$
D) $\frac { 8 x ^ { 8 } + 1 } { \left( x ^ { 8 } + x - 4 \right) }$
E) $\frac { 8 x ^ { 7 } + 1 } { ( \ln 3 ) \left( x ^ { 8 } + x - 4 \right) }$

Question

Quizplus · Accepted Answer

To find $y'$, we use the chain rule and the derivative of the logarithm function. The derivative of $\log_3(u)$ with respect to $u$ is $\frac{1}{u \ln(3)}$, and the derivative of $u = 4 - x - x^8$ with respect to $x$ is $-1 - 8x^7$. Combining these, we get $\frac{-1 - 8x^7}{(4 - x - x^8) \ln(3)}$, which simplifies to $\frac{8x^7 + 1}{(x^8 + x - 4) \ln(3)}$.

Find y′y ^ { \prime }y′ y=log⁡3(4−x−x8)y = \log _ { 3 } \left( 4 - x - x ^ { 8 } \right)y=log3​(4−x−x8)

Find $y ^ { \prime }$ $y = \log _ { 3 } \left( 4 - x - x ^ { 8 } \right)$