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${:[f(3)=-1" and "f^(')(3)=0],[g(x)=x^(3)]:} Evaluate (d)/(dx)[f(x)*g(x)] at x=3.$

$\begin{array}{l} f(3)=-1 \text { and } f^{\prime}(3)=0 \\ g(x)=x^{3} \end{array}$ $\newline$ Evaluate $\frac{d}{d x}[f(x) \cdot g(x)]$ at $x=3$ .

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Find derivatives of sine and cosine functions

Full solution

Q.

\begin{array}{l} f(3)=-1 \text { and } f^{\prime}(3)=0 \\ g(x)=x^{3} \end{array}

\newline

Evaluate

\frac{d}{d x}[f(x) \cdot g(x)]

at

x=3

.

Identify Rule: Identify the rule for differentiating a product of two functions. $\newline$ Product Rule: $(\frac{d}{dx})[u(x)\cdot v(x)] = u'(x)\cdot v(x) + u(x)\cdot v'(x)$
Differentiate Functions: Differentiate $f(x)$ and $g(x)$ separately. $f'(x)$ is given as $0$ and $g'(x)$ is the derivative of $x^3$ , which is $3x^2$ .
Apply Product Rule: Plug in the derivatives and the original functions into the Product Rule. $\newline$ $(\frac{d}{dx})[f(x)*g(x)] = f'(x)*g(x) + f(x)*g'(x)$ $\newline$ $(\frac{d}{dx})[f(x)*g(x)] = 0*x^3 + (-1)*3x^2$
Evaluate at $x=3$ : Evaluate the expression at $x=3$ .
$\frac{d}{dx}[f(x)*g(x)]$ at $x=3$ = $0\cdot3^3 + (-1)\cdot3\cdot3^2$
$\frac{d}{dx}[f(x)*g(x)]$ at $x=3$ = $0 + (-1)\cdot3\cdot9$
Simplify Final Answer: Simplify the expression to find the final answer. $\newline$ $(\frac{d}{dx})[f(x)*g(x)]$ at $x=3$ = $0 - 27$ $\newline$ $(\frac{d}{dx})[f(x)*g(x)]$ at $x=3$ = $-27$

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