If f(x) = $\frac { 1 } { x ^ { 3 } + x ^ { 2 } }$ , find $f ^ { \prime }$ (x).
A) $\frac { 1 } { \left( x ^ { 3 } + x ^ { 2 } \right) ^ { 2 } }$
B) $\frac { x ^ { 3 } + x ^ { 2 } } { \left( x ^ { 3 } + x ^ { 2 } \right) ^ { 2 } }$
C) $\frac { - 3 x ^ { 2 } - 2 x } { \left( x ^ { 3 } + x ^ { 2 } \right) ^ { 2 } }$
D) - $\frac { 2 x + 1 } { x ^ { 2 } ( x + 1 ) ^ { 2 } }$

Question

Quizplus · Accepted Answer

To find $f'(x)$, we use the quotient rule for differentiation, $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$, where $f(x) = \frac{g(x)}{h(x)}$, $g(x) = 1$, and $h(x) = x^3 + x^2$. Differentiating $g(x)$ and $h(x)$ gives $g'(x) = 0$ and $h'(x) = 3x^2 + 2x$. Substituting these into the quotient rule formula yields $f'(x) = \frac{- (3x^2 + 2x)}{(x^3 + x^2)^2}$, which matches choice C.

If F(x) = 1x3+x2\frac { 1 } { x ^ { 3 } + x ^ { 2 } }x3+x21​

If F(x) = $\frac { 1 } { x ^ { 3 } + x ^ { 2 } }$