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.

Identify Product Rule: question_prompt: How much is (d)/(dx)[f(x)*h(x)] when x=-2? Find Derivative of f(x)*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 (d)/(dx)[u*v] = u' * v + u * 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)*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)=(1)/(x), so h' (x) = -1/(x^2). Apply Product Rule at x=-2: Now we need to evaluate h'(-2). h'(-2) = -1/((-2)^2) = -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*(1/(-2)) + (-8)*(-1/4). This simplifies to -2 - (-2), which equals 0.

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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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Grade 8

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)]

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 $0$ .