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Let
f be a function such that
f(-2)=-8 and
f^(')(-2)=4.
Let
h be the function
h(x)=(1)/(x).

Evaluate
(d)/(dx)[f(x)*h(x)] at
x=-2.

- Let $f$ be a function such that $f(-2)=-8$ and $f^{\prime}(-2)=4$ . $\newline$ - Let $h$ be the function $h(x)=\frac{1}{x}$ . $\newline$ Evaluate $\frac{d}{d x}[f(x) \cdot h(x)]$ at $x=-2$ .

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Write variable expressions for arithmetic sequences

Full solution

Q. - Let

f

be a function such that

f(-2)=-8

and

f^{\prime}(-2)=4

.

\newline

- Let

h

be the function

h(x)=\frac{1}{x}

.

\newline

Evaluate

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

at

x=-2

.

Identify Product Rule: question_prompt: How much is $(\frac{d}{dx}[f(x)*h(x)])$ when $x=-2$ ?
Find Derivative of $f(x)\cdot h(x)$ : We know that to find the derivative of a product of two functions, we use the product rule. The product rule states that $\frac{d}{dx}[u\cdot v] = u' \cdot v + u \cdot v'$ , where $u$ and $v$ are functions of $x$ , and $u'$ and $v'$ are their respective derivatives.
Evaluate $h'(-2)$ : Let's find the derivative of $f(x)\cdot h(x)$ using the product rule. We have $u=f(x)$ and $v=h(x)$ . We know $f(-2)=-8$ and $f'(-2)=4$ . We also know $h(x)=\frac{1}{x}$ , so $h'(x) = -\frac{1}{x^2}$ .
Apply Product Rule at $x=-2$ : Now we need to evaluate $h'(-2)$ . $h'(-2) = -\frac{1}{((-2)^2)} = -\frac{1}{4}$ .
Plug in Values: Using the product rule, $(d)/(dx)[f(x)*h(x)]$ at $x=-2$ is $f^{'}(-2)*h(-2) + f(-2)*h^{'}(-2)$ .
Simplify and Calculate: We plug in the values we know: $4\times\left(\frac{1}{-2}\right) + (-8)\times\left(-\frac{1}{4}\right)$ .
Simplify and Calculate: We plug in the values we know: $4*(1/(-2)) + (-8)*(-1/4)$ . This simplifies to $-2 - (-2)$ , which equals $0$ .

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