Solving For Dy/dx And Tangent Line Of -20x³ + 8x⁴⁰y + Y² = -11

Understanding Implicit Differentiation

In the realm of calculus, particularly when dealing with equations where y is not explicitly defined as a function of x, we turn to a technique called implicit differentiation. This powerful tool allows us to find the derivative dy/dx even when y is intertwined with x in a complex manner. Our mission here is to find dy/dx for the equation -20x³ + 8x⁴⁰y + y² = -11 and then determine the equation of the tangent line at the point (1, 1). This involves a step-by-step application of differentiation rules and algebraic manipulation.

First, the main approach to implicit differentiation lies in differentiating both sides of the equation with respect to x, while keeping in mind that y is a function of x. This means that whenever we encounter a term involving y, we must apply the chain rule. The chain rule is essential when differentiating composite functions, which are functions within functions. In our case, the chain rule dictates that the derivative of y² with respect to x is 2y(dy/dx). This is a crucial step in implicit differentiation, as it correctly accounts for the dependency of y on x.

Now, applying this principle to our equation, we differentiate each term with respect to x. The derivative of -20x³ is -60x², a straightforward application of the power rule. The term 8x⁴⁰y, however, requires the product rule since it is a product of two functions of x: 8x⁴⁰ and y. The product rule states that the derivative of uv is u'v + uv', where u' and v' are the derivatives of u and v, respectively. Applying this rule, we get (320x³⁹ * y) + (8x⁴⁰ * dy/dx). Lastly, the derivative of y² with respect to x is 2y(dy/dx), as we discussed earlier, and the derivative of the constant -11 is simply 0. This initial differentiation step transforms our original equation into a form where dy/dx is present and can be isolated.

Calculating dy/dx

Now, with the derivatives calculated, our equation looks like this: -60x² + 320x³⁹y + 8x⁴⁰(dy/dx) + 2y(dy/dx) = 0. The next step is to isolate dy/dx. This involves gathering all terms containing dy/dx on one side of the equation and moving all other terms to the other side. Doing so, we have 8x⁴⁰(dy/dx) + 2y(dy/dx) = 60x² - 320x³⁹y. The common factor dy/dx can then be factored out from the left side, giving us (dy/dx)(8x⁴⁰ + 2y) = 60x² - 320x³⁹y. Finally, to solve for dy/dx, we divide both sides of the equation by the factor (8x⁴⁰ + 2y), resulting in dy/dx = (60x² - 320x³⁹y) / (8x⁴⁰ + 2y). This is our general expression for the derivative dy/dx for the given equation. It tells us the slope of the tangent line to the curve at any point (x, y) that satisfies the original equation. This expression is a key result, but to find the tangent line at a specific point, we need to evaluate it at that point.

With the general expression for dy/dx in hand, the next step is to evaluate it at the point (1, 1). This will give us the slope of the tangent line to the curve at this particular point. Substituting x = 1 and y = 1 into our expression for dy/dx, we get dy/dx = (60(1)² - 320(1)³⁹(1)) / (8(1)⁴⁰ + 2(1)). This simplifies to (60 - 320) / (8 + 2), which further simplifies to -260 / 10, giving us dy/dx = -26. This value, -26, represents the slope of the tangent line at the point (1, 1). This is a crucial piece of information needed to determine the equation of the tangent line.

Determining the Tangent Line Equation

Now that we have the slope of the tangent line at (1, 1), which is -26, we can determine the equation of the tangent line. The point-slope form of a line is given by y - y₁ = m(x - x₁), where m is the slope of the line and (x₁, y₁) is a point on the line. In our case, m = -26 and (x₁, y₁) = (1, 1). Substituting these values into the point-slope form, we get y - 1 = -26(x - 1). To express the equation in slope-intercept form (y = mx + b), we need to distribute and rearrange the equation. Distributing the -26, we have y - 1 = -26x + 26. Adding 1 to both sides, we get y = -26x + 27. This is the equation of the tangent line to the curve at the point (1, 1), expressed in slope-intercept form. This result gives us a clear linear approximation of the curve's behavior near the point (1, 1).

In conclusion, by employing the technique of implicit differentiation, we successfully found the derivative dy/dx for the equation -20x³ + 8x⁴⁰y + y² = -11. We then evaluated this derivative at the point (1, 1) to find the slope of the tangent line at that point. Finally, using the point-slope form of a line, we determined the equation of the tangent line to be y = -26x + 27. This exercise demonstrates the power and versatility of implicit differentiation in handling equations where variables are intertwined and provides a solid understanding of how to find tangent lines to implicitly defined curves.

Summary of Steps

To recap, here are the key steps we took to solve this problem:

1. Implicit Differentiation: We differentiated both sides of the equation -20x³ + 8x⁴⁰y + y² = -11 with respect to x, using the product rule and chain rule as necessary.
2. Isolating dy/dx: We rearranged the resulting equation to isolate dy/dx, obtaining a general expression for the derivative.
3. Evaluating at (1, 1): We substituted x = 1 and y = 1 into the expression for dy/dx to find the slope of the tangent line at the point (1, 1).
4. Tangent Line Equation: We used the point-slope form of a line to determine the equation of the tangent line, then converted it to slope-intercept form.

By following these steps, we were able to successfully find both dy/dx and the equation of the tangent line, highlighting the importance of implicit differentiation in calculus.

Problem

Given the equation below, find $\frac{d y}{d x}$.

$-20 x^3+8 x^{40} y+y^2=-11$

$\frac{d y}{d x} = $

Now, find the equation of the tangent line to the curve at $(1,1)$. Write your answer in $m x+b$ format.

$y=$

Solution

Finding dy/dx using Implicit Differentiation

To find $\frac{dy}{dx}$ for the equation $-20x^3 + 8x^{40}y + y^2 = -11$, we'll use implicit differentiation. This means we will differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule where necessary.

Differentiating both sides with respect to $x$:

$\frac{d}{dx}(-20x^3) + \frac{d}{dx}(8x^{40}y) + \frac{d}{dx}(y^2) = \frac{d}{dx}(-11)$

Now, let's differentiate each term:

1. $\frac{d}{dx}(-20x^3) = -60x^2$
2. For $\frac{d}{dx}(8x^{40}y)$, we'll use the product rule, which states that $(uv)' = u'v + uv'$. Here, let $u = 8x^{40}$ and $v = y$. So, $u' = 320x^{39}$ and $v' = \frac{dy}{dx}$. Thus, $\frac{d}{dx}(8x^{40}y) = 320x^{39}y + 8x^{40}\frac{dy}{dx}$
3. For $\frac{d}{dx}(y^2)$, we'll use the chain rule. The derivative of $y^2$ with respect to $y$ is $2y$, and then we multiply by $\frac{dy}{dx}$. So, $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$
4. $\frac{d}{dx}(-11) = 0$ (since the derivative of a constant is 0)

Putting it all together, we have:

$-60x^2 + 320x^{39}y + 8x^{40}\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

Now, we need to solve for $\frac{dy}{dx}$. First, group the terms with $\frac{dy}{dx}$ on one side and the other terms on the other side:

$8x^{40}\frac{dy}{dx} + 2y\frac{dy}{dx} = 60x^2 - 320x^{39}y$

Factor out $\frac{dy}{dx}$:

$\frac{dy}{dx}(8x^{40} + 2y) = 60x^2 - 320x^{39}y$

Now, divide by $(8x^{40} + 2y)$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{60x^2 - 320x^{39}y}{8x^{40} + 2y}$

We can simplify this expression by dividing both the numerator and the denominator by 2:

$\frac{dy}{dx} = \frac{30x^2 - 160x^{39}y}{4x^{40} + y}$

Finding the Tangent Line Equation at (1, 1)

To find the equation of the tangent line at the point $(1, 1)$, we need to find the slope of the tangent line at this point. We'll do this by plugging $x = 1$ and $y = 1$ into our expression for $\frac{dy}{dx}$:

$\frac{dy}{dx}|_{(1,1)} = \frac{30(1)^2 - 160(1)^{39}(1)}{4(1)^{40} + 1} = \frac{30 - 160}{4 + 1} = \frac{-130}{5} = -26$

So, the slope of the tangent line at $(1, 1)$ is $-26$.

Now, we'll use the point-slope form of a line, which is $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point. In our case, $m = -26$ and $(x_1, y_1) = (1, 1)$. Plugging these values in, we get:

$y - 1 = -26(x - 1)$

Now, let's convert this to slope-intercept form ($y = mx + b$):

$y - 1 = -26x + 26$

Add 1 to both sides:

$y = -26x + 27$

So, the equation of the tangent line to the curve at $(1, 1)$ is $y = -26x + 27$.

$\frac{dy}{dx} = \frac{30x^2 - 160x^{39}y}{4x^{40} + y}$

$y = -26x + 27$