Let f(x)=(3x^(2)+1)/(x+2). Find f^(')(-3). Choose 1 answer: (A) 10 (B) 28 (C) -28 (D) -10

Quotient Rule Application: To find the derivative of the function f(x) = (3x^2 + 1) / (x + 2), we will use the quotient rule which states that if f(x) = g(x) / h(x), then f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2. Here, g(x) = 3x^2 + 1 and h(x) = x + 2.Derivative of g(x): First, we find the derivative of g(x) = 3x^2 + 1. Using the power rule, we get g'(x) = d/dx(3x^2) + d/dx(1) = 6x + 0 = 6x.Derivative of h(x): Next, we find the derivative of h(x) = x + 2. The derivative of a constant is 0, and the derivative of x with respect to x is 1, so h'(x) = d/dx(x) + d/dx(2) = 1 + 0 = 1.Applying Quotient Rule: Now we apply the quotient rule: f'(x) = (6x * (x + 2) - (3x^2 + 1) * 1) / (x + 2)^2.Simplifying Numerator: We simplify the numerator: 6x * (x + 2) - (3x^2 + 1) = 6x^2 + 12x - 3x^2 - 1 = 3x^2 + 12x - 1.Evaluate at x = -3: Now we have f'(x) = (3x^2 + 12x - 1) / (x + 2)^2. We need to evaluate this derivative at x = -3.Calculate Numerator: Substitute x = -3 into the derivative: f'(-3) = (3(-3)^2 + 12(-3) - 1) / (-3 + 2)^2.Calculate Denominator: Calculate the numerator: 3(-3)^2 + 12(-3) - 1 = 3(9) - 36 - 1 = 27 - 36 - 1 = -10.Final Result: Calculate the denominator: (-3 + 2)^2 = (-1)^2 = 1.Now we have f'(-3) = (-10) / 1 = -10.

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Let
f(x)=(3x^(2)+1)/(x+2).
Find
f^(')(-3).
Choose 1 answer:
(A)
10
(B) 28
(C) -28
(D) -10

Let $f(x)=\frac{3 x^{2}+1}{x+2}$ . $\newline$ Find $f^{\prime}(-3)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\mathbf{1 0}$ $\newline$ (B) $28$ $\newline$ (C) $-28$ $\newline$ (D) $-10$

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Calculus

Find derivatives of using multiple formulae

Full solution

Q. Let

f(x)=\frac{3 x^{2}+1}{x+2}

\newline

Find

f^{\prime}(-3)

\newline

Choose

1

answer:

\newline

(A)

\mathbf{1 0}

\newline

(B)

28

\newline

(C)

-28

\newline

(D)

-10

Quotient Rule Application: To find the derivative of the function $f(x) = \frac{3x^2 + 1}{x + 2}$ , we will use the quotient rule which states that if $f(x) = \frac{g(x)}{h(x)}$ , then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$ . Here, $g(x) = 3x^2 + 1$ and $h(x) = x + 2$ .
Derivative of $g(x)$ : First, we find the derivative of $g(x) = 3x^2 + 1$ . Using the power rule, we get $g'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(1) = 6x + 0 = 6x$ .
Derivative of $h(x)$ : Next, we find the derivative of $h(x) = x + 2$ . The derivative of a constant is $0$ , and the derivative of $x$ with respect to $x$ is $1$ , so $h'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(2) = 1 + 0 = 1$ .
Applying Quotient Rule: Now we apply the quotient rule: $f'(x) = \frac{(6x \cdot (x + 2) - (3x^2 + 1) \cdot 1)}{(x + 2)^2}$ .
Simplifying Numerator: We simplify the numerator: $6x \times (x + 2) - (3x^2 + 1) = 6x^2 + 12x - 3x^2 - 1 = 3x^2 + 12x - 1$ .
Evaluate at $x = -3$ : Now we have $f'(x) = \frac{3x^2 + 12x - 1}{(x + 2)^2}$ . We need to evaluate this derivative at $x = -3$ .
Calculate Numerator: Substitute $x = -3$ into the derivative: $f'(-3) = \frac{3(-3)^2 + 12(-3) - 1}{(-3 + 2)^2}$ .
Calculate Denominator: Calculate the numerator: $3(-3)^2 + 12(-3) - 1 = 3(9) - 36 - 1 = 27 - 36 - 1 = -10$ .
Final Result: Calculate the denominator: $(-3 + 2)^2 = (-1)^2 = 1$ .
Final Result: Calculate the denominator: $(-3 + 2)^2 = (-1)^2 = 1$ .Now we have $f'(-3) = (-10) / 1 = -10$ .