Questions: Find the equation of the tangent of the curve (2 x+y ln x=4 y) at the point (left(1, frac12right)). a. (8 y=5 x+1) b. (8 y=-5 x-1) c. (8 y=5 x-1) d. (8 y=-5 x+1)

$Find the equation of the tangent of the curve (2 x+y ln x=4 y) at the point (left(1, frac12right)). a. (8 y=5 x+1) b. (8 y=-5 x-1) c. (8 y=5 x-1) d. (8 y=-5 x+1)$

Transcript text: Question 9 Not yet answered Marked out of 1.00 Flag question Find the equation of the tangent of the curve $2 x+y \ln x=4 y$ at the point $\left(1, \frac{1}{2}\right)$. a. $\quad 8 y=5 x+1$ b. $8 y=-5 x-1$ c. $8 y=5 x-1$ d. $8 y=-5 x+1$

Solution

Solution Steps

Step 1: Rewrite the equation in a differentiable form

The given equation is: \[ 2x + y \ln x = 4y. \] We rewrite it as: \[ 2x + y \ln x - 4y = 0. \]

Step 2: Differentiate implicitly with respect to $x$

To find the slope of the tangent line, we differentiate both sides of the equation with respect to $x$. Using implicit differentiation: \[ \frac{d}{dx}(2x) + \frac{d}{dx}(y \ln x) - \frac{d}{dx}(4y) = 0. \] This gives: \[ 2 + \left(\frac{dy}{dx} \ln x + y \cdot \frac{1}{x}\right) - 4 \frac{dy}{dx} = 0. \]

Step 3: Solve for $\frac{dy}{dx}$

Rearrange the terms to solve for $\frac{dy}{dx}$: \[ 2 + \frac{dy}{dx} \ln x + \frac{y}{x} - 4 \frac{dy}{dx} = 0. \] Combine like terms: \[ 2 + \frac{y}{x} + \frac{dy}{dx} (\ln x - 4) = 0. \] Solve for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = -\frac{2 + \frac{y}{x}}{\ln x - 4}. \]

Step 4: Evaluate $\frac{dy}{dx}$ at the point $\left(1, \frac{1}{2}\right)$

Substitute $x = 1$ and $y = \frac{1}{2}$ into the expression for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = -\frac{2 + \frac{\frac{1}{2}}{1}}{\ln 1 - 4}. \] Simplify: \[ \frac{dy}{dx} = -\frac{2 + \frac{1}{2}}{0 - 4} = -\frac{\frac{5}{2}}{-4} = \frac{5}{8}. \] The slope of the tangent line at the point $\left(1, \frac{1}{2}\right)$ is $\frac{5}{8}$.

Step 5: Write the equation of the tangent line

Using the point-slope form of a line: \[ y - y_1 = m(x - x_1), \] where $m = \frac{5}{8}$ and $(x_1, y_1) = \left(1, \frac{1}{2}\right)$, we get: \[ y - \frac{1}{2} = \frac{5}{8}(x - 1). \] Multiply both sides by 8 to eliminate the fraction: \[ 8y - 4 = 5x - 5. \] Rearrange to match the given options: \[ 8y = 5x - 1. \]