Problem 301 - Calculus Volume 1

Question

For the following exercises, use implicit differentiation to find \(\frac{d y}{d x}\) $$ 6 x^{2}+3 y^{2}=12 $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{-2x}{y}\)

Step 1: Differentiate the Equation

First, differentiate both sides of the equation with respect to \(x\). Remember that \(y\) is a function of \(x\), so you'll use the chain rule for terms involving \(y\). For \(6x^2\), the derivative is \(12x\). For \(3y^2\), apply the chain rule: the derivative is \(6y \cdot \frac{dy}{dx}\). The derivative of a constant is zero. So, the differentiated equation is \(12x + 6y \cdot \frac{dy}{dx} = 0\).

Step 2: Isolate the Derivative Term

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\). Subtract \(12x\) from both sides to get \(6y \cdot \frac{dy}{dx} = -12x\).

Step 3: Solve for \(\frac{dy}{dx}\)

To isolate \(\frac{dy}{dx}\), divide both sides of the equation by \(6y\). This gives \(\frac{dy}{dx} = \frac{-12x}{6y}\). Simplify the expression: \(\frac{dy}{dx} = \frac{-2x}{y}\).

Key Concepts

chain rule

The chain rule is a fundamental tool in calculus useful when differentiating composite functions. In our exercise, we deal with the equation \(6x^2 + 3y^2 = 12\), where \(y\) is a function of \(x\). This raises the need for implicit differentiation, which is where the chain rule comes into play. To apply the chain rule:

• First, recognize expressions where differentiation of \(y\) (a function of \(x\)) is needed. For \(3y^2\), the derivative isn't straightforward because \(y\) is treated as a dependent variable.
• Use the chain rule by first differentiating the outer function, \(2y\), yielding \(6y\). Then, multiply by the derivative of the inner function, \(\frac{dy}{dx}\), leading to \(6y \cdot \frac{dy}{dx}\).

This approach captures how changes in \(x\) affect \(y\) indirectly through \(x\), a key characteristic of implicit differentiation. By employing the chain rule here, we appropriately account for the relationship between \(x\) and \(y\), ensuring these dependencies are considered in our calculations.

derivatives

Derivatives measure the rate at which a function changes concerning its variables, which in calculus, signifies the slope of the function at any point. When dealing with implicit differentiation in our exercise, derivatives of different terms within the equation must be carefully observed and applied. Consider the expression \(6x^2 + 3y^2 = 12\):

• When taking the derivative of \(6x^2\) with respect to \(x\), you simply follow standard differentiation rules, resulting in \(12x\).
• The term \(3y^2\) is more complex because \(y\) is dependent on \(x\). With implicit differentiation, apply the chain rule to gain the derivative \(6y \cdot \frac{dy}{dx}\).
• For a constant like \(12\), the derivative equals 0 since constants do not change.

Note that derivatives allow one to turn geometric concepts of slopes and tangents into analytical expressions, providing insights into behavioral dynamics of functions across variables.

equations

Equations in calculus often tie together multiple variables, requiring a systematic approach to solve or analyze them efficiently. In our exercise, the equation \(6x^2 + 3y^2 = 12\) is a relation between \(x\) and \(y\). It describes an ellipse graphically, though without explicit information on how \(y\) changes with \(x\).To solve for \(\frac{dy}{dx}\), the slope of \(y\) concerning \(x\) using implicit differentiation, perform the following steps:

• Differentiate the entire equation concerning \(x\).
• You now have an equation \(12x + 6y \cdot \frac{dy}{dx} = 0\), capturing all variables and their derivatives.
• Isolate \(\frac{dy}{dx}\) by first subtracting \(12x\) from both sides, then dividing through by the coefficient of \(\frac{dy}{dx}\), which is \(6y\).

This yields \(\frac{dy}{dx} = \frac{-2x}{y}\), illustrating the proportional change of \(y\) in response to \(x\) alterations. This quotient gives a dynamic insight into the behavior of the original geometric representation, reflecting shifts in \(x\) relative to \(y\).