What is the value of (d)/(dx)((x^(2)-2x+3)/(x+1)) at x=1 ? Choose 1 answer: (A) -2 (B) -(1)/(2) (C) 1 (D) -1

Apply Quotient Rule: Apply the quotient rule to find the derivative of the function (x^(2)-2x+3)/(x+1). The quotient rule is given by (d/dx)(u/v) = (v(u') - u(v')) / v^2, where u = x^2 - 2x + 3 and v = x + 1.Differentiate u: Differentiate u = x^2 - 2x + 3 with respect to x. u' = (d/dx)(x^2) - (d/dx)(2x) + (d/dx)(3) u' = 2x - 2 + 0 u' = 2x - 2Differentiate v: Differentiate v = x + 1 with respect to x. v' = (d/dx)(x) + (d/dx)(1) v' = 1 + 0 v' = 1Substitute into Quotient Rule: Substitute u, u', v, and v' into the quotient rule formula. (d/dx)((x^2 - 2x + 3)/(x + 1)) = ((x + 1)(2x - 2) - (x^2 - 2x + 3)(1)) / (x + 1)^2Simplify Expression: Simplify the expression obtained in Step 4. (d/dx)((x^2 - 2x + 3)/(x + 1)) = (2x^2 - 2x + 2 - x^2 + 2x - 3) / (x + 1)^2 (d/dx)((x^2 - 2x + 3)/(x + 1)) = (x^2 - 1) / (x + 1)^2Evaluate at x=1: Evaluate the derivative at x=1. (d/dx)((x^2 - 2x + 3)/(x + 1))|_{x=1} = (1^2 - 1) / (1 + 1)^2 (d/dx)((x^2 - 2x + 3)/(x + 1))|_{x=1} = (1 - 1) / (2)^2 (d/dx)((x^2 - 2x + 3)/(x + 1))|_{x=1} = 0 / 4 (d/dx)((x^2 - 2x + 3)/(x + 1))|_{x=1} = 0

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What is the value of
(d)/(dx)((x^(2)-2x+3)/(x+1)) at
x=1 ?
Choose 1 answer:
(A) -2
(B)
-(1)/(2)
(C) 1
(D) -1

What is the value of $\frac{d}{d x}\left(\frac{x^{2}-2 x+3}{x+1}\right)$ at $x=1$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-2$ $\newline$ (B) $-\frac{1}{2}$ $\newline$ (C) $1$ $\newline$ (D) $-1$

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Calculus

Find derivatives of using multiple formulae

Full solution

Q. What is the value of

\frac{d}{d x}\left(\frac{x^{2}-2 x+3}{x+1}\right)

x=1

\newline

Choose

1

answer:

\newline

(A)

-2

\newline

(B)

-\frac{1}{2}

\newline

(C)

1

\newline

(D)

-1

Apply Quotient Rule: Apply the quotient rule to find the derivative of the function $(x^{2}-2x+3)/(x+1)$ . $\newline$ The quotient rule is given by $(d/dx)(u/v) = (v(u') - u(v')) / v^{2}$ , where $u = x^{2} - 2x + 3$ and $v = x + 1$ .
Differentiate $u$ : Differentiate $u = x^2 - 2x + 3$ with respect to $x$ .
$u' = \frac{d}{dx}(x^2) - \frac{d}{dx}(2x) + \frac{d}{dx}(3)$
$u' = 2x - 2 + 0$
$u' = 2x - 2$
Differentiate $v$ : Differentiate $v = x + 1$ with respect to $x$ .
$v' = \frac{d}{dx}(x) + \frac{d}{dx}(1)$
$v' = 1 + 0$
$v' = 1$
Substitute into Quotient Rule: Substitute $u$ , $u'$ , $v$ , and $v'$ into the quotient rule formula. $(\frac{d}{dx})\left(\frac{x^2 - 2x + 3}{x + 1}\right) = \frac{(x + 1)(2x - 2) - (x^2 - 2x + 3)(1)}{(x + 1)^2}$
Simplify Expression: Simplify the expression obtained in Step $4$ . $\newline$ $(\frac{d}{dx})\left(\frac{x^2 - 2x + 3}{x + 1}\right) = \frac{2x^2 - 2x + 2 - x^2 + 2x - 3}{(x + 1)^2}$ $\newline$ $(\frac{d}{dx})\left(\frac{x^2 - 2x + 3}{x + 1}\right) = \frac{x^2 - 1}{(x + 1)^2}$
Evaluate at $x=1$ : Evaluate the derivative at $x=1$ .
$\frac{d}{dx}\left(\frac{x^2 - 2x + 3}{x + 1}\right)\bigg|_{x=1} = \frac{1^2 - 2\cdot 1 + 3}{(1 + 1)^2}$
$\frac{d}{dx}\left(\frac{x^2 - 2x + 3}{x + 1}\right)\bigg|_{x=1} = \frac{1 - 2 + 3}{2^2}$
$\frac{d}{dx}\left(\frac{x^2 - 2x + 3}{x + 1}\right)\bigg|_{x=1} = \frac{2}{4}$
$\frac{d}{dx}\left(\frac{x^2 - 2x + 3}{x + 1}\right)\bigg|_{x=1} = \frac{1}{2}$