Question

Find \(d y / d x,\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}.\)

Short answer

Answer: \(\frac{dy}{dx} = \frac{16xy^2 - 2x(x^2+y^2+x)-(x^2+y^2)(2x+1)}{2y(x^2+y^2+x)+8y^2-2y(x^2+y^2)}\)

Step-by-step solution

Differentiate both sides of the equation with respect to x

We are given the equation \((x^2+y^2)(x^2+y^2+x)=8xy^2\). We need to differentiate both sides with respect to \(x\) using the product rule and implicit differentiation. For implicit differentiation, whenever we differentiate a term including \(y\) with respect to \(x\), we need to multiply by \(\frac{dy}{dx}\). Using the product rule for differentiation, we have: \([(x^2+y^2)'(x^2+y^2+x)+(x^2+y^2)(x^2+y^2+x)']=16xy^2+8y^2\frac{dy}{dx}\).

Differentiate the individual terms

Now we differentiate the terms involving \(x\) and \(y.\) letting U = x^2+y^2: \(\frac{dU}{dx} = 2x + 2y\frac{dy}{dx}\) \(\frac{d(x^2+y^2+x)}{dx} = 2x + 1 + 2y\frac{dy}{dx}.\) Substituting the derivatives back into the equation: \((2x+2y\frac{dy}{dx})(x^2+y^2+x)+(x^2+y^2)(2x+1+2y\frac{dy}{dx})=16xy^2+8y^2\frac{dy}{dx}\)

Expand and group together the dy/dx terms

Expand the terms and group all the terms involving \(\frac{dy}{dx}\) together: \(2x(x^2+y^2+x)+2y\frac{dy}{dx}(x^2+y^2+x)+(x^2+y^2)(2x+1)+2y(x^2+y^2)\frac{dy}{dx}=16xy^2+8y^2\frac{dy}{dx}\) Combine the \(\frac{dy}{dx}\) terms: \(\frac{dy}{dx}(2y(x^2+y^2+x)+8y^2-2y(x^2+y^2)) = 16xy^2 - 2x(x^2+y^2+x) - (x^2+y^2)(2x+1)\)

Divide by the coefficient of dy/dx to solve for dy/dx

Divide both sides of the equation by the coefficient of \(\frac{dy}{dx}\) to get the final expression for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{16xy^2 - 2x(x^2+y^2+x)-(x^2+y^2)(2x+1)}{2y(x^2+y^2+x)+8y^2-2y(x^2+y^2)}\)

Key concepts

Product Rule

The product rule is an essential concept in differentiation. It helps us find the derivative of a product of two functions. For example, if we have two functions, \(u(x)\) and \(v(x)\), the product rule states that the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

In other words, you differentiate the first function and multiply it by the second one. Then, add it to the product of the first function and the derivative of the second function.

In our exercise, the expression \((x^2+y^2)(x^2+y^2+x)\) represents the product of two functions of \(x\). When using the product rule here, we differentiate each term with respect to \(x\) and handle any \(y\) terms by applying implicit differentiation. This is necessary because \(y\) is dependent on \(x\), and hence, its derivative includes \(\frac{dy}{dx}\).

Applying the product rule to this particular problem requires careful attention to detail, especially when substituting the differentiated components back into the expanded equation. This allows us to combine like terms and solve for \(\frac{dy}{dx}\).

Differentiation

Differentiation is a key tool in calculus, used to calculate how a function changes as its input changes. In simpler terms, it helps us find the rate of change or slope at any given point on a curve. When differentiating an expression, we usually aim to find \(\frac{dy}{dx}\), which denotes the derivative of \(y\) with respect to \(x\).

In our exercise, we start with an equation involving both \(x\) and \(y\). Because \(y\) is implicitly defined in relation to \(x\), implicit differentiation becomes necessary. Implicit differentiation involves differentiating both sides of an equation, taking care to apply the product and chain rules where appropriate.

When dealing with both \(x\) and \(y\) terms, keep in mind:

• Differentiating terms in \(x\) directly as you normally would.
• For terms in \(y\), every differentiation results in an additional \(\frac{dy}{dx}\) factor.

These steps allow you to handle more complex expressions and solve for \(\frac{dy}{dx}\) as performed in the given solution.

Calculus

Calculus is the mathematical study of continuous change. It is divided mainly into two parts: differentiation and integration. Differentiation, the focus of this article, stems from understanding how functions change. It's essential in many fields, including physics, engineering, economics, and biology.

The exercise showcases a real-world application of calculus in differentiation. Given the complexity of the equation, we apply calculus tools to derive a formula for \(\frac{dy}{dx}\). This gives insight into how changes in \(x\) affect \(y\), which is crucial in dynamic systems where variables are interdependent.

To solve equations like the one provided, calculus offers:

• Product and chain rules for handling products and compositions of functions.
• Implicit differentiation for dealing with equations that define \(y\) implicitly in terms of \(x\).
• Approaches for isolating terms and simplifying expressions to find derivatives.

This highlights the versatility and power of calculus in understanding and modeling change.