Questions: Differentiate. y=(3x-1)/(6x^2+1) (-18x^2+9x+4)/(6x^2+1)^2 (18x^3-36x^2+15x)/(6x^2+1)^2 (-18x^2+12x+3)/(6x^2+1)^2 (54x^2-12x+3)/(6x^2+1)^2

Transcript text: Homework 6 Started: Nov 1 at 9pm Quiz Instructions 12 Question 34 Differentiate. \[ y=\frac{3 x-1}{6 x^{2}+1} \] $\frac{-18 x^{2}+9 x+4}{\left(6 x^{2}+1\right)^{2}}$ $\frac{18 x^{3}-36 x^{2}+15 x}{\left(6 x^{2}+1\right)^{2}}$ $\frac{-18 x^{2}+12 x+3}{\left(6 x^{2}+1\right)^{2}}$ $\frac{54 x^{2}-12 x+3}{\left(6 x^{2}+1\right)^{2}}$ Previous

Solution

Solution Steps

To differentiate the given function $ y = \frac{3x - 1}{6x^2 + 1} $, we will use the quotient rule. The quotient rule states that if you have a function $ y = \frac{u}{v} $, then its derivative $ y' $ is given by $ y' = \frac{u'v - uv'}{v^2} $, where $ u' $ and $ v' $ are the derivatives of $ u $ and $ v $ respectively. Here, $ u = 3x - 1 $ and $ v = 6x^2 + 1 $.

Step 1: Define the Functions

Let $ u = 3x - 1 $ and $ v = 6x^2 + 1 $.

Step 2: Compute the Derivatives

The derivatives of $ u $ and $ v $ are calculated as follows: \[ u' = \frac{du}{dx} = 3 \] \[ v' = \frac{dv}{dx} = 12x \]

Step 3: Apply the Quotient Rule

Using the quotient rule, the derivative $ y' $ is given by: \[ y' = \frac{u'v - uv'}{v^2} \] Substituting the values we have: \[ y' = \frac{3(6x^2 + 1) - (3x - 1)(12x)}{(6x^2 + 1)^2} \]

Step 4: Simplify the Expression

Now, we simplify the expression: \[ y' = \frac{18x^2 + 3 - (36x^2 - 12x)}{(6x^2 + 1)^2} \] This simplifies to: \[ y' = \frac{18x^2 + 3 - 36x^2 + 12x}{(6x^2 + 1)^2} = \frac{-18x^2 + 12x + 3}{(6x^2 + 1)^2} \]