Let f be a function such that f(1)=0 and f^(')(1)=-7. Let g be the function g(x)=sqrtx. Evaluate (d)/(dx)[(f(x))/(g(x))] at x=1

Apply Quotient Rule: Apply the quotient rule to find the derivative of (f(x)/g(x)). The quotient rule states that if you have a function h(x) = u(x)/v(x), then h'(x) = (u'(x)v(x) - u(x)v'(x))/(v(x))^2. Here, u(x) = f(x) and v(x) = g(x) = sqrt(x).Differentiate f(x) and g(x: Differentiate f(x) and g(x) with respect to x. We are given that f'(1) = -7, but we do not have a general expression for f'(x). However, we only need its value at x=1. To differentiate g(x) = sqrt(x), we use the power rule. The derivative of x^(1/2) is (1/2)x^(-1/2). So, g'(x) = (1/2)x^(-1/2).Evaluate Derivatives at x=1: Evaluate the derivatives at x=1. We have f'(1) = -7 and g'(1) = (1/2)(1)^(-1/2) = 1/2.Plug Values into Formula: Plug the values into the quotient rule formula. Using the values from Step 3, we get: h'(1) = (f'(1)g(1) - f(1)g'(1))/(g(1))^2. We know that f(1) = 0 and g(1) = sqrt(1) = 1. So, h'(1) = (-7 * 1 - 0 * (1/2)) / (1)^2.Simplify Expression: Simplify the expression to find the derivative at x=1. h'(1) = (-7 - 0) / 1. h'(1) = -7.

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Let
f be a function such that
f(1)=0 and
f^(')(1)=-7.
Let
g be the function
g(x)=sqrtx.

Evaluate
(d)/(dx)[(f(x))/(g(x))] at
x=1

- Let $f$ be a function such that $f(1)=0$ and $f^{\prime}(1)=-7$ . $\newline$ - Let $g$ be the function $g(x)=\sqrt{x}$ . $\newline$ Evaluate $\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]$ at $x=1$ .

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Q. - Let

f

be a function such that

f(1)=0

and

f^{\prime}(1)=-7

\newline

- Let

g

be the function

g(x)=\sqrt{x}

\newline

Evaluate

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

x=1

Apply Quotient Rule: Apply the quotient rule to find the derivative of $\frac{f(x)}{g(x)}$ . The quotient rule states that if you have a function $h(x) = \frac{u(x)}{v(x)}$ , then $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ . Here, $u(x) = f(x)$ and $v(x) = g(x) = \sqrt{x}$ .
Differentiate $f(x)$ and $g(x)$ : Differentiate $f(x)$ and $g(x)$ with respect to $x$ . We are given that $f'(1) = -7$ , but we do not have a general expression for $f'(x)$ . However, we only need its value at $x=1$ . To differentiate $g(x) = \sqrt{x}$ , we use the power rule. The derivative of $x^{1/2}$ is $g(x)$ $0$ . So, $g(x)$ $1$ .
Evaluate Derivatives at $x=1$ : Evaluate the derivatives at $x=1$ . We have $f'(1) = -7$ and $g'(1) = \left(\frac{1}{2}\right)(1)^{-\frac{1}{2}} = \frac{1}{2}$ .
Plug Values into Formula: Plug the values into the quotient rule formula. $\newline$ Using the values from Step $3$ , we get: $\newline$ $h'(1) = \frac{f'(1)g(1) - f(1)g'(1)}{(g(1))^2}$ . $\newline$ We know that $f(1) = 0$ and $g(1) = \sqrt{1} = 1$ . $\newline$ So, $h'(1) = \frac{(-7 \times 1 - 0 \times (\frac{1}{2}))}{(1)^2}$ .
Simplify Expression: Simplify the expression to find the derivative at $x=1$ . $h'(1) = (\$-7$ - $0$ ) / $1$ .\) $h'(1) = -7.$