Let $$ f(x)=\frac{4}{\left(3 x^{2}+3 x-6\right)^{3}} $$. Find $$ f^{\prime}(x) $$ $$ f^{\prime}(x)=\square $$

Question

UpStudy AI · Accepted Answer

To find the derivative $$ f'(x) $$ of the function

$$
f(x) = \frac{4}{(3x^2 + 3x - 6)^3},
$$

we can use the quotient rule or the chain rule. Here, we will use the chain rule, rewriting the function as:

$$
f(x) = 4(3x^2 + 3x - 6)^{-3}.
$$

Now, we will differentiate $$ f(x) $$ using the chain rule. The derivative of $$ g(x) = (h(x))^{-3} $$ is given by:

$$
g'(x) = -3(h(x))^{-4} h'(x),
$$

where $$ h(x) = 3x^2 + 3x - 6 $$.

First, we need to find $$ h'(x) $$:

$$
h(x) = 3x^2 + 3x - 6,
$$
$$
h'(x) = 6x + 3.
$$

Now we can apply the chain rule to find $$ f'(x) $$:

$$
f'(x) = 4 \cdot (-3)(3x^2 + 3x - 6)^{-4} \cdot h'(x).
$$

Substituting $$ h'(x) $$:

$$
f'(x) = 4 \cdot (-3)(3x^2 + 3x - 6)^{-4} \cdot (6x + 3).
$$

Simplifying this expression:

$$
f'(x) = -12(6x + 3)(3x^2 + 3x - 6)^{-4}.
$$

Thus, the final result for the derivative $$ f'(x) $$ is:

$$
f'(x) = -\frac{12(6x + 3)}{(3x^2 + 3x - 6)^4}.
$$

So, we can write:

$$
f'(x) = -\frac{12(6x + 3)}{(3x^2 + 3x - 6)^4}.
$$
$$ f'(x) = -\frac{12(6x + 3)}{(3x^2 + 3x - 6)^4} $$.

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