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Let
f be a function such that
f(9)=-54 and
f^(')(9)=6.
Let
h be the function
h(x)=sqrtx.

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

- Let $f$ be a function such that $f(9)=-54$ and $f^{\prime}(9)=6$ . $\newline$ - Let $h$ be the function $h(x)=\sqrt{x}$ . $\newline$ Evaluate $\frac{d}{d x}\left[\frac{f(x)}{h(x)}\right]$ at $x=9$ .

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Transformations of absolute value functions

Full solution

Q. - Let

f

be a function such that

f(9)=-54

and

f^{\prime}(9)=6

.

\newline

- Let

h

be the function

h(x)=\sqrt{x}

.

\newline

Evaluate

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

at

x=9

.

Quotient Rule Derivative: To find the derivative of the function $\frac{f(x)}{h(x)}$ at $x=9$ , we will use the quotient rule for derivatives. The quotient rule states that if we have a function $g(x) = \frac{u(x)}{v(x)}$ , then its derivative $g'(x)$ is given by $g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ . Here, $u(x) = f(x)$ and $v(x) = h(x) = \sqrt{x}$ .
Derivative of $f(x)$ : First, we need to find the derivative of $u(x) = f(x)$ . We are given that $f'(9) = 6$ , which means that the derivative of $f(x)$ at $x=9$ is $6$ .
Derivative of sqrt(x): Next, we need to find the derivative of $v(x) = h(x) = \sqrt{x}$ . The derivative of $\sqrt{x}$ with respect to $x$ is $(1/2)x^{(-1/2)}$ . So, $v'(x) = (1/2)x^{(-1/2)}$ .
Evaluate $v'(x)$ at $x=9$ : Now we will evaluate $v'(x)$ at $x=9$ . Substituting $x$ with $9$ , we get $v'(9) = (1/2)9^{(-1/2)} = (1/2)(1/\sqrt{9}) = (1/2)(1/3) = 1/6$ .
Values of $u(9)$ and $v(9)$ : We also need the values of $u(9)$ and $v(9)$ to use in the quotient rule formula. We are given that $f(9) = -54$ , so $u(9) = -54$ . We also have $v(9) = h(9) = \sqrt{9} = 3$ .
Apply Quotient Rule: Now we can apply the quotient rule. We have all the necessary values: $\newline$ $u'(9) = 6$ , $\newline$ $v(9) = 3$ , $\newline$ $u(9) = -54$ , $\newline$ $v'(9) = \frac{1}{6}$ . $\newline$ Substituting these into the quotient rule formula, we get: $\newline$ $g'(9) = \frac{u'(9)v(9) - u(9)v'(9)}{(v(9))^2}$ $\newline$ $g'(9) = \frac{6 \times 3 - (-54) \times (\frac{1}{6})}{(3)^2}$
Simplify the Expression: Simplify the expression: $\newline$ $g'(9) = \frac{18 + 9}{9}$ $\newline$ $g'(9) = \frac{27}{9}$ $\newline$ $g'(9) = 3$

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