Differentiate.
-f(x) = $\frac { 2 x - \left( x ^ { 2 } + 1 \right) ^ { 7 } } { 3 }$
A) f'(x) = $f ^ { \prime } ( x ) = 2 - 7 \left( x ^ { 2 } + 1 \right) ^ { 6 }$
B) f'(x) = $\frac { 2 } { 3 }$ - $\frac { 14 } { 3 }$ x( $x ^ { 2 }$ + 1 $)^6$
C) f'(x) = $\frac { 2 - x } { 3 \left( x ^ { 2 } + 1 \right) ^ { 6 } }$
D) none of these

Question

Quizplus · Accepted Answer

To differentiate $f(x) = \frac { 2x - (x^2 + 1)^7 } { 3 }$, apply the quotient rule and the chain rule. The derivative of $2x$ is $2$, and for $-(x^2 + 1)^7$, apply the chain rule: $7(x^2 + 1)^6 \cdot 2x$. Combine these results and divide by 3 to match the original fraction, resulting in $\frac { 2 } { 3 }$ - $\frac { 14 } { 3 }$x($x^2 + 1)^6$.

Differentiate 2x−(x2+1)73\frac { 2 x - \left( x ^ { 2 } + 1 \right) ^ { 7 } } { 3 }32x−(x2+1)7​

Differentiate $\frac { 2 x - \left( x ^ { 2 } + 1 \right) ^ { 7 } } { 3 }$