Question

Find \(d y / d x\) by implicit differentiation. \(x^{3} y^{3}-y=x\)

Short answer

The derivative \(y'\) or quick answer is given by \(-\frac{3x^{2}y^{3} - 1}{1 - 3x^{3}y^{2}}\).

Step-by-step solution

Differentiate Both Sides of the Equation

Differentiate both sides of the equation with respect to \(x\). Remember to apply the derivative rule for a product \(uv\): \((uv)' = u'v + uv'\). \nHence, differentiating \(x^{3}y^{3}\) gives \(3x^{2}y^{3} + x^{3}(3y^{2}y')\), where \(y'\) refers to \(dy/dx\). \nDifferentiating \(-y\) with respect to \(x\) gives \(-y'\). Differentiating on the right side of the equation \(x\) gives \(1\).

Rearrange the Equation

Rearrange the equation to isolate \(y'\), the derivative of \(y\) with respect to \(x\). After differentiating both sides, you get the equation \(3x^{2}y^{3} + 3x^{3}y^{2}y' - y' = 1\). This equation can be rearranged to isolate \(y'\), yielding \(y' = \frac{1 - 3x^{2}y^{3}}{3x^{3}y^{2} - 1}\).

Simplify the Result

The derivative \(y'\) obtained can be simplified by dividing numerator and denominator by \(-1\). This gives the final answer for \(y'\) which is \(-\frac{3x^{2}y^{3} - 1}{1 - 3x^{3}y^{2}}\).

Key concepts

Derivative of Implicit Functions

Understanding the concept of taking derivatives of implicit functions is crucial for solving many problems in calculus. An implicit function is one where the variables involved are not separated and the function is not explicitly solved for one variable in terms of another. The classic example is an equation like \(x^3y^3 - y = x\), which does not explicitly solve for \(y\) in terms of \(x\) or vice versa.

To find the derivative of such a function with respect to \(x\), or \(\frac{dy}{dx}\), we apply implicit differentiation. This involves differentiating both sides of the equation with respect to \(x\) and treating \(y\) as a function of \(x\). Wherever \(y\) appears, we multiply by \(\frac{dy}{dx}\) (or \(y'\)) since it's a function of \(x\). After differentiation, we then solve for \(\frac{dy}{dx}\), rearranging the equation if necessary. This approach allows us to find the rate of change of \(y\) with respect to \(x\) even when the function's form is complex or intertwined.

Product Rule of Differentiation

When differentiating products of two functions, the product rule is essential. The rule comes into play when you have an equation where two functions are multiplied together, such as \(u\) and \(v\). According to the product rule, the derivative of the product \(uv\) is \(u'v + uv'\), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively.

For instance, in the implicit differentiation of \(x^3y^3\), we treat \(x^3\) as \(u\) and \(y^3\) as \(v\). Differentiating \(u\) with respect to \(x\) gives us \(3x^2\) while differentiating \(v\), since it's a function of \(y\), requires us to also multiply by the derivative of \(y\) with respect to \(x\), resulting in \(3y^2\frac{dy}{dx}\). Combining these, we have the derivative of the product as \(3x^2y^3 + x^3(3y^2)\frac{dy}{dx}\). The product rule simplifies the differentiation process of complex functions by breaking them down into manageable parts.

Rearranging Equations for Differentiation

Once you've differentiated an implicit function with variables \(x\) and \(y\), you might need to 'rearrange' the equation to solve for \(\frac{dy}{dx}\) or \(y'\). This process involves collecting all the terms containing \(y'\) on one side and moving the remaining terms to the other side.

Using algebraic operations, we can then factor out \(y'\) and divide across to isolate it. For example, after applying the product rule and differentiating both sides of our equation, we get an expression like \(3x^2y^3 + 3x^3y^2y' - y' = 1\). We then gather the \(y'\) terms to one side resulting in \(3x^3y^2y' - y' = 1 - 3x^2y^3\), factor out \(y'\), and then divide to solve for it: \(y' = \frac{1 - 3x^2y^3}{3x^3y^2 - 1}\). This step is crucial because it turns an implicit relationship into an explicit formula for the derivative, which is necessary for further analysis or computations involving the derivative.

Calculus of a Single Variable

The field of calculus often involves dealing with a single variable, typically denoted \(x\), and its associated functions. When we talk about calculus of a single variable, we are primarily concerned with differentiation and integration within the context of functions that are defined with respect to one independent variable.

In the context of implicit differentiation, even though it may seem like we are dealing with two variables (\(x\) and \(y\)), we are actually considering \(y\) as an implicit function of the single variable \(x\). This is important for understanding how changes in \(x\) affect \(y\), and vice versa. By focusing on \(x\) as the independent variable, we can use the rules and techniques of single variable calculus, such as the product rule and implicit differentiation, to find the derivatives of more complex, implicitly defined functions. Recognizing that \(y\) is dependent on \(x\) allows us to apply these concepts effectively and derive the required solutions for problems presented in single variable calculus.