Problem 303 - Calculus Volume 1

Question

Use implicit differentiation to find \(\frac{d y}{d x}\). $$ 3 x^{3}+9 x y^{2}=5 x^{3} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{2x^2 - 3y^2}{6xy}\)

Step 1: Differentiate Both Sides Respect to x

Start by differentiating both sides of the equation with respect to \(x\). The original equation is \(3x^3 + 9xy^2 = 5x^3\). Differentiating the left side involves using the product rule on \(9xy^2\). The derivative of \(3x^3\) is \(9x^2\) and for \(9xy^2\) it is \(9(y^2 + 2xy\frac{dy}{dx})\). Differentiating the right side \(5x^3\) gives \(15x^2\).

Step 2: Simplify and Solve for \(\frac{dy}{dx}\)

Substitute the derivatives found in Step 1 into the equation: \(9x^2 + 9(y^2 + 2xy\frac{dy}{dx}) = 15x^2\). Simplify this to: \(9x^2 + 9y^2 + 18xy\frac{dy}{dx} = 15x^2\). Rearrange to isolate terms with \(\frac{dy}{dx}\): \(18xy\frac{dy}{dx} = 15x^2 - 9x^2 - 9y^2\).

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

Subtract and simplify on the right: \(15x^2 - 9x^2 - 9y^2 = 6x^2 - 9y^2\). The equation becomes \(18xy\frac{dy}{dx} = 6x^2 - 9y^2\). Divide both sides by \(18xy\) to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{6x^2 - 9y^2}{18xy}\). Simplify the expression further, divide the numerator and denominator by 3: \(\frac{dy}{dx} = \frac{2x^2 - 3y^2}{6xy}\).

Key Concepts

calculus

Calculus is a branch of mathematics focused on studying change and motion. It is primarily divided into two branches: differential calculus and integral calculus. In the context of this exercise, we emphasize differential calculus, particularly implicit differentiation, to find derivatives.

Implicit differentiation is an extension of regular differentiation. It allows us to handle situations where equations are not easily solved for one variable in terms of another, like in the given exercise equation: \[3x^3 + 9xy^2 = 5x^3\].

• Implicit differentiation applies the derivative rules to both sides of the equation with respect to a chosen variable, often \(x\).
• This technique is especially useful because some equations do not lend themselves to straightforward solutions where solving for \(y\) and differentiating directly is feasible.

After differentiating, terms involving derivatives of \(y\) with respect to \(x\) are often found, such as \(\frac{dy}{dx}\), which can be solved using algebraic methods afterward. Implicit differentiation gives a powerful tool for calculus solving challenges that might not be possible with direct differentiation approaches.

derivatives

Derivatives represent the rate at which a function is changing at any given point, making them a cornerstone of calculus. Mathematically speaking, if you have a function \(f(x)\), the derivative, denoted \(f'(x)\) or \(\frac{df}{dx}\), provides the slope of the tangent line to the function at any point \(x\).

In the exercise, derivatives are used to determine how changes in \(x\) impact changes in \(y\). This is crucial for implicit differentiation:

• The process involves differentiating both sides of an equation with respect to \(x\).
• When \(y\) is dependent on \(x\), the chain rule introduces \(\frac{dy}{dx}\) in the derivative calculation.

Understanding how to differentiate correctly, especially when dealing with products of terms or powers of terms like \(3x^3\) or \(9xy^2\), is foundational. As shown in the step-by-step solution, the derivative of \(9xy^2\) involves both the product and chain rules, producing additional terms like \(2xy\frac{dy}{dx}\). The ability to effectively manage these derivatives is essential to correctly applying implicit differentiation and algebra to isolate the desired derivative.

product rule

The product rule is an essential differentiation rule applied when differentiating expressions that are products of two functions. If you have two functions, \(u(x)\) and \(v(x)\), their product \(uv\) has a derivative given by:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]

• This rule ensures that we effectively account for the rate of change of each function within the product.
• It is applicable in scenarios similar to those seen in the exercise with terms like \(9xy^2\), which is a product of \(9x\) and \(y^2\).

Applying the product rule allows us to break down the derivative of \(9xy^2\) into two components: the derivative of \(9x\) times \(y^2\), plus \(9x\) times the derivative of \(y^2\). Since \(y\) is a function of \(x\), the derivation of \(y^2\) involves the chain rule, introducing an extra \(2y\frac{dy}{dx}\) term.

Mastery of the product rule is crucial for successfully tackling implicit differentiation problems, as seen in the original exercise, allowing the decomposition of complex terms into manageable parts for further algebraic manipulation.