Let g be a function such that g(9)=0 and g^(')(9)=2. Let h be the function h(x)=sqrtx. Evaluate (d)/(dx)[g(x)*h(x)] at x=9.

Product Rule Explanation: We need to use the product rule for differentiation, which is (f * g)' = f' * g + f * g'.Derivative of h(x): First, let's find the derivative of h(x) = sqrt(x). The derivative h'(x) is 1/(2sqrt(x)).Calculate h'(9): Now we plug in x=9 into h'(x) to get h'(9) = 1/(2sqrt(9)) = 1/6.Given Values for g: We know g(9)=0 and g'(9)=2 from the problem statement.Product Rule Application: Using the product rule: (g(x) * h(x))' at x=9 is g'(9) * h(9) + g(9) * h'(9).Substitute Values: Substitute the known values: 2 * sqrt(9) + 0 * (1/6).Simplify Result: This simplifies to 2 * 3 + 0 = 6.

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Let
g be a function such that
g(9)=0 and
g^(')(9)=2.
Let
h be the function
h(x)=sqrtx.

Evaluate
(d)/(dx)[g(x)*h(x)] at
x=9.

- Let $g$ be a function such that $g(9)=0$ and $g^{\prime}(9)=2$ . $\newline$ - Let $h$ be the function $h(x)=\sqrt{x}$ . $\newline$ Evaluate $\frac{d}{d x}[g(x) \cdot h(x)]$ at $x=9$ .

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Grade 8

Write variable expressions for arithmetic sequences

Full solution

Q. - Let

g

be a function such that

g(9)=0

and

g^{\prime}(9)=2

\newline

- Let

h

be the function

h(x)=\sqrt{x}

\newline

Evaluate

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

x=9

Product Rule Explanation: We need to use the product rule for differentiation, which is $(f \cdot g)' = f' \cdot g + f \cdot g'$ .
Derivative of $h(x)$ : First, let's find the derivative of $h(x) = \sqrt{x}$ . The derivative $h'(x)$ is $\frac{1}{2\sqrt{x}}$ .
Calculate $h'(9)$ : Now we plug in $x=9$ into $h'(x)$ to get $h'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}$ .
Given Values for $g$ : We know $g(9)=0$ and $g'(9)=2$ from the problem statement.
Product Rule Application: Using the product rule: $(g(x) \cdot h(x))'$ at $x=9$ is $g'(9) \cdot h(9) + g(9) \cdot h'(9)$ .
Substitute Values: Substitute the known values: $2 \times \sqrt{9} + 0 \times \left(\frac{1}{6}\right)$ .
Simplify Result: This simplifies to $2 \times 3 + 0 = 6$ .