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Let
g be a function such that
g(1)=-2 and
g^(')(1)=7.
Let
h be the function
h(x)=sqrtx.

Let
F be a function defined as
F(x)=g(x)*h(x).

F^(')(1)=

- Let $g$ be a function such that $g(1)=-2$ and $g^{\prime}(1)=7$ . $\newline$ - Let $h$ be the function $h(x)=\sqrt{x}$ . $\newline$ Let $F$ be a function defined as $F(x)=g(x) \cdot h(x)$ . $\newline$ $F^{\prime}(1)=$

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Evaluate exponential functions

Full solution

Q. - Let

g

be a function such that

g(1)=-2

and

g^{\prime}(1)=7

.

\newline

- Let

h

be the function

h(x)=\sqrt{x}

.

\newline

Let

F

be a function defined as

F(x)=g(x) \cdot h(x)

.

\newline

F^{\prime}(1)=

Apply Product Rule: Use the product rule for differentiation, which states that $(f*g)' = f'*g + f*g'$ .
Differentiate $h(x)$ : Differentiate $h(x) = \sqrt{x}$ to get $h'(x)$ . $h'(x) = \frac{1}{2\sqrt{x}}$ .
Evaluate $h'(1)$ : Evaluate $h'(1)$ by substituting $x$ with $1$ . $h'(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2}$ .
Use Product Rule Formula: Now, use the values $g(1)=-2$ , $g'(1)=7$ , and $h'(1)=\frac{1}{2}$ in the product rule formula.
Apply Product Rule: Apply the product rule: $F'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)$ .
Substitute $x$ in $F'(x)$ : Substitute $x$ with $1$ to find $F'(1)$ : $F'(1) = g'(1)\cdot h(1) + g(1)\cdot h'(1)$ .
Plug in Known Values: Plug in the known values: $F'(1) = 7\sqrt{1} + (-2)\left(\frac{1}{2}\right)$ .
Calculate $F'(1)$ : Calculate the values: $F'(1) = 7\cdot1 + (-2)\cdot(\frac{1}{2}) = 7 - 1$ .
Finish Calculation: Finish the calculation: $F'(1) = 7 - 1 = 6$ .

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