Question

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

Short answer

Question: Find the derivative \(\frac{dy}{dx}\) of the function (x²+y²)(x²+y²+x) = 8xy². Answer: \(\frac{dy}{dx} = \frac{8y^{2}-2xy^{2}-2y^{3}-x^{3}}{4x^{2}+4xy^{2}+3xy^{2}}\)

Step-by-step solution

Differentiate implicitly with respect to x

We have \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\). To differentiate this equation with respect to x, we'll need to apply the product rule \((u v)^{'}=u^{'} v+u v^{'}\) and the chain rule \(\left(f(g(x))\right)^{'}=f^{'}(g(x))\cdot g^{'}(x)\). Differentiate implicitly with respect to x: \(\frac{d}{d x}\left[\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)\right]=\frac{d}{d x}\left[8 x y^{2}\right]\)

Apply the product rule and chain rule

Apply the product rule, chain rule, and differentiate: \((2 x + 2 y \frac{d y}{d x})(x^{2} + y^{2} + x) +(x^{2}+y^{2})(2 x+2 y\frac{d y}{d x}+1) = 8y^{2}+16x y\frac{d y}{d x}\)

Solve for \(\frac{d y}{d x}\)

Collect all terms involving \(\frac{d y}{d x}\) on one side of the equation and factor out \(\frac{d y}{d x}\): \(4x^{2}\frac{dy}{dx}+4xy^{2}\frac{dy}{dx}+3xy^{2}\frac{dy}{dx}=(8y^{2}-2xy^{2}-2y^{3}-x^{3})\) Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(4x^{2}+4xy^{2}+3xy^{2})=8y^{2}-2xy^{2}-2y^{3}-x^{3}\) Now, divide by the term within the parentheses to find \(\frac{dy}{dx}\): \(\frac{dy}{dx}=\frac{8y^{2}-2xy^{2}-2y^{3}-x^{3}}{4x^{2}+4xy^{2}+3xy^{2}}\) Thus, the derivative \(\frac{d y}{d x}\) is given by: \(\frac{d y}{d x}=\frac{8y^{2}-2xy^{2}-2y^{3}-x^{3}}{4x^{2}+4xy^{2}+3xy^{2}}\)

Key concepts

Product Rule

The product rule is an essential tool in calculus, especially when dealing with expressions that involve products of two functions. When you have a function that is a product of two subfunctions, like \(u(x) v(x)\), the product rule helps you find the derivative. The rule states:

• \((uv)' = u'v + uv'\)

This means that to differentiate the product of two functions, you take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function.
This technique is particularly useful when dealing with complex expressions like \( (x^2 + y^2)(x^2 + y^2 + x) \). In such cases, each expression can behave as its own function, making the application of the product rule necessary to differentiate with respect to \(x\).
Using these steps ensures you account for all parts of the expression and allow the implicit component (with respect to \(y\)) to remain efficiently handled.

Chain Rule

The chain rule is a powerful and quite necessary tool when dealing with composite functions. This rule is applicable in situations where one function is nested inside another. Mathematically, if you have a composite function like \(f(g(x))\), the chain rule lets you differentiate it as follows:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

This process involves two steps: first, differentiate the outer function (while keeping the inner function as is), and then multiply the result by the derivative of the inner function.
In cases where the inner function involves \(y\), an implicit variable, you would also need its derivative with respect to \(x\). This application is evident in our exercise where terms like \(y^2\) require the chain rule to differentiate implicating \(2y \frac{dy}{dx}\).
The chain rule helps manage these layers of composition, ensuring that the differentiation process addresses all levels of nested functions accurately.

Differentiation with Respect to x

Differentiation, by definition, involves finding the rate at which a function changes with respect to another variable. In our context, we're differentiating with respect to \(x\), which means analyzing how a function changes as \(x\) changes.
This method applies both explicit and implicit parts of the function. For expressions involving \(y\), treating \(y\) as a function of \(x\), called implicit differentiation, becomes vital. For any term involving \(y\), you tack on an \(\frac{dy}{dx}\) as part of its chain of differentiation.
This approach allows mathematicians and students alike to find derivatives even for complex expressions intertwining \(x\) and \(y\). In our example exercise, differentiating the compound expressions with respect to \(x\) means finding derivatives of each term, bearing in mind whether they involve a simple \(x\), a complex composition, or an implicit involvement of \(y\).
This systematic differentiation assures complete and correct derivation across all terms, leading smoothly to the solution where every component involving changes in \(x\) has been addressed.