Let y=(1-2x)/(3x^(2)). What is the value of (dy)/(dx) at x=1 ? Choose 1 answer: (A) 0 (B) -(1)/(3) (C) 1 (D) (1)/(9)

Identify function: Identify the function to differentiate. We are given the function y=(1-2x)/(3x^(2)). We need to find its derivative with respect to x, which is denoted as (dy)/(dx).Use quotient rule: Use the quotient rule to differentiate the function. The quotient rule states that if we have a function that is the quotient of two functions, u(x)/v(x), then its derivative is given by (v(x)u'(x) - u(x)v'(x))/(v(x))^2. Here, u(x) = 1 - 2x and v(x) = 3x^2.Differentiate u and v: Differentiate u(x) and v(x) with respect to x. The derivative of u(x) = 1 - 2x is u'(x) = -2. The derivative of v(x) = 3x^2 is v'(x) = 6x.Apply quotient rule: Apply the quotient rule. (dy)/(dx) = ((3x^2)(-2) - (1 - 2x)(6x)) / (3x^2)^2 (dy)/(dx) = (-6x^2 - (6x - 12x^2)) / 9x^4 (dy)/(dx) = (-6x^2 - 6x + 12x^2) / 9x^4 (dy)/(dx) = (6x^2 - 6x) / 9x^4Simplify expression: Simplify the expression. (dy)/(dx) = (6x(x - 1)) / (9x^4) (dy)/(dx) = (6(x - 1)) / (9x^3) (dy)/(dx) = 2(x - 1) / (3x^3)Evaluate at x=1: Evaluate the derivative at x=1. (dy)/(dx) at x=1 is 2(1 - 1) / (3(1)^3) (dy)/(dx) at x=1 is 2(0) / 3 (dy)/(dx) at x=1 is 0

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Math Problems

Grade 8

Write variable expressions for arithmetic sequences

Full solution

Q. Let

y=\frac{1-2 x}{3 x^{2}}

\newline

What is the value of

\frac{d y}{d x}

x=1

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B)

-\frac{1}{3}

\newline

(C)

1

\newline

(D)

\frac{1}{9}

Identify function: Identify the function to differentiate. $\newline$ We are given the function $y=\frac{1-2x}{3x^{2}}$ . We need to find its derivative with respect to $x$ , which is denoted as $\frac{dy}{dx}$ .
Use quotient rule: Use the quotient rule to differentiate the function. $\newline$ The quotient rule states that if we have a function that is the quotient of two functions, $\frac{u(x)}{v(x)}$ , then its derivative is given by $\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$ . Here, $u(x) = 1 - 2x$ and $v(x) = 3x^2$ .
Differentiate $u$ and $v$ : Differentiate $u(x)$ and $v(x)$ with respect to $x$ . The derivative of $u(x) = 1 - 2x$ is $u'(x) = -2$ . The derivative of $v(x) = 3x^2$ is $v'(x) = 6x$ .
Apply quotient rule: Apply the quotient rule. $\newline$ $(dy)/(dx) = ((3x^2)(-2) - (1 - 2x)(6x)) / (3x^2)^2$ $\newline$ $(dy)/(dx) = (-6x^2 - (6x - 12x^2)) / 9x^4$ $\newline$ $(dy)/(dx) = (-6x^2 - 6x + 12x^2) / 9x^4$ $\newline$ $(dy)/(dx) = (6x^2 - 6x) / 9x^4$
Simplify expression: Simplify the expression. $\newline$ $(dy)/(dx) = (6x(x - 1)) / (9x^4)$ $\newline$ $(dy)/(dx) = (6(x - 1)) / (9x^3)$ $\newline$ $(dy)/(dx) = 2(x - 1) / (3x^3)$
Evaluate at $x=1$ : Evaluate the derivative at $x=1$ . $\frac{dy}{dx}$ at $x=1$ is $\frac{2(1 - 1)}{3(1)^3}$ $\frac{dy}{dx}$ at $x=1$ is $\frac{2(0)}{3}$ $\frac{dy}{dx}$ at $x=1$ is $x=1$ $0$