Questions: Given the function (4 x^2 y^2+5 x y=26) : a. Differentiate implicitly with respect to (x). Your answer should be the first line in the implicit differentiation process. Do not simplify or factor terms. b. Find the slope of the line tangent to the graph of the function at the point ((2,1)). c. Find the equation of the tangent line at the point ((2,1)) in the form (y=mx+b). [ y= ]

Solution Steps

To solve the given problems, we will follow these steps:

a. Implicit Differentiation: Differentiate the given equation implicitly with respect to $x$. This involves applying the chain rule and product rule where necessary.

b. Slope of the Tangent Line: Use the result from implicit differentiation to find the derivative $\frac{dy}{dx}$ and evaluate it at the point $(2, 1)$.

c. Equation of the Tangent Line: Use the point-slope form of a line equation, $y - y_1 = m(x - x_1)$, where $m$ is the slope found in part b, and $(x_1, y_1)$ is the point $(2, 1)$.

Given the function:

$$4x^2y^2 + 5xy = 26$$

We need to differentiate implicitly with respect to $x$. Using the product rule and chain rule, we differentiate each term:

1. Differentiate $4x^2y^2$:

   • Use the product rule: $u = 4x^2$ and $v = y^2$.
   • $\frac{d}{dx}(uv) = u'v + uv'$.
   • $u' = 8x$ and $v' = 2y \frac{dy}{dx}$.
   • So, $\frac{d}{dx}(4x^2y^2) = 8xy^2 + 4x^2(2y \frac{dy}{dx}) = 8xy^2 + 8x^2y \frac{dy}{dx}$.

2. Differentiate $5xy$:

   • Use the product rule: $u = 5x$ and $v = y$.
   • $u' = 5$ and $v' = \frac{dy}{dx}$.
   • So, $\frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx}$.

3. Differentiate the constant $26$:

   • The derivative of a constant is $0$.

Putting it all together, the implicit differentiation of the function is:

$$8xy^2 + 8x^2y \frac{dy}{dx} + 5y + 5x \frac{dy}{dx} = 0$$

To find the slope of the tangent line at the point $(2, 1)$, substitute $x = 2$ and $y = 1$ into the differentiated equation:

$$8(2)(1)^2 + 8(2)^2(1) \frac{dy}{dx} + 5(1) + 5(2) \frac{dy}{dx} = 0$$

Simplify:

$$16 + 32 \frac{dy}{dx} + 5 + 10 \frac{dy}{dx} = 0$$

Combine like terms:

$$21 + 42 \frac{dy}{dx} = 0$$

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

$$42 \frac{dy}{dx} = -21$$

$$\frac{dy}{dx} = -\frac{21}{42} = -\frac{1}{2}$$

The slope of the tangent line is $-\frac{1}{2}$.

The equation of the tangent line is in the form $y = mx + b$, where $m$ is the slope and $(x_1, y_1) = (2, 1)$ is the point of tangency.

Using the point-slope form of a line:

$$y - y_1 = m(x - x_1)$$

Substitute $m = -\frac{1}{2}$, $x_1 = 2$, and $y_1 = 1$:

$$y - 1 = -\frac{1}{2}(x - 2)$$

Simplify:

$$y - 1 = -\frac{1}{2}x + 1$$

$$y = -\frac{1}{2}x + 2$$

Final Answer