Question

Suppose that \(x y+y^{3}=2 .\) Then implicit differentiation twice with respect to \(x\) yields in turn: (a) \(x y^{\prime}+y+3 y^{2} y^{\prime}=0\) (b) \(x y^{\prime \prime}+y^{\prime}+y^{\prime}+3 y^{2} y^{\prime \prime}+6 y\left(y^{\prime}\right)^{2}=0 .\)

Short answer

(a) \(x y' + y + 3y^2 y' = 0\); (b) \(x y'' + 2y' + 6y(y')^2 + 3y^2 y'' = 0\).

Step-by-step solution

Differentiate Implicitly for First Derivative

Given the function \(x y + y^3 = 2\), our goal is to determine the first derivative using implicit differentiation. Differentiate both sides with respect to \(x\). Apply the product rule to the term \(xy\), resulting in \(x \cdot y' + y\). For the term \(y^3\), use the chain rule to get \(3y^2 y'\). Set the derivative of the constant 2 to 0, as the derivative of a constant is zero. Combine these results to obtain the equation: \[ x y' + y + 3y^2 y' = 0. \]

Differentiate Again for Second Derivative

Now, differentiate the entire equation derived in step 1, \(x y' + y + 3y^2 y' = 0\), with respect to \(x\). Start by applying the product rule and chain rule as needed. The first term \(x y'\) differentiates to \(x y'' + y'\), the second term \(y\) differentiates to \(y'\), and the third term \(3y^2 y'\) requires the product rule: \[ \frac{d}{dx} (3y^2 y') = 3(2yy' y' + y^2 y'') = 6y(y')^2 + 3y^2 y''. \]

Combine and Simplify

Combine all differentiated terms:\[ x y'' + y' + y' + 6y(y')^2 + 3y^2 y'' = 0. \]Simplify by combining like terms, particularly \(y' + y'\) which becomes \(2y'\). Therefore, the second derivative equation becomes:\[ x y'' + 2y' + 6y(y')^2 + 3y^2 y'' = 0. \]This step concludes the implicit differentiation twice.

Key concepts

Derivatives

When studying calculus, derivatives are essential for understanding how a function changes. But what exactly is a derivative? Simply put, a derivative describes the rate of change of a function with respect to a variable. In practice, it tells us how much the function's output value changes as the input changes. For example, if you consider the function of a car's speed over time, the derivative would give us the car's acceleration.

In mathematical notation, if you have a function \( f(x) \), then the derivative of this function is represented as \( f'(x) \) or \( \frac{df}{dx} \). Derivatives are foundational for many concepts in calculus, including implicit differentiation, which is often used when dealing with equations where one variable is not explicitly solved in terms of the other.

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. It allows us to differentiate functions that are composed of other functions, making it essential in many differentiation problems. When you have a function inside another function, you use the chain rule to differentiate.

Consider a situation where you have a function \( f(g(x)) \). To differentiate this, the chain rule states that you take the derivative of the outer function \( f \) with respect to \( g \), and multiply it by the derivative of the inner function \( g \) with respect to \( x \). That is, \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

In our original problem, applying the chain rule was necessary to differentiate terms like \( y^3 \). We first differentiated the outer function \( y^3 \) with respect to \( y \) to get \( 3y^2 \), and then multiplied by \( \frac{dy}{dx} \) because \( y \) itself is a function of \( x \).

Product Rule

The product rule is a handy tool when you need to differentiate products of functions. So, how does it work? If you have two functions, say \( u(x) \) and \( v(x) \), their product is \( u(x) \cdot v(x) \). The product rule says: to find the derivative of this product, you should differentiate one function at a time while keeping the other fixed, then add the results.

Mathematically, this is expressed as \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Each part involves differentiating one of the functions and multiplying it by the other function as-is.

In the given exercise, the term \( xy \) was differentiated using the product rule. First, we took the derivative of \( x \) and multiplied it by \( y \), and then the derivative of \( y \) was multiplied by \( x \). This step is crucial because it helps in dealing with more complex relationships between variables in an equation.

Implicit Function

An implicit function is an equation that defines a relationship between variables in a way where the dependent variable is not isolated, contrary to a function like \( y = f(x) \). In the real world, many relationships are naturally implicit. For instance, the equation of a circle \( x^2 + y^2 = r^2 \) connects \( x \) and \( y \) without solving for one in terms of the other.

Differentiating implicit functions requires implicit differentiation. This technique is used when the function isn't explicitly given in terms of one variable. By taking the derivative of both sides with respect to \( x \), we implicitly differentiate, which might involve techniques like the chain and product rules as applied in the given exercise.

Implicit differentiation is particularly useful because it allows us to find higher order derivatives, like the second derivative as shown in our solution. This is crucial when analyzing the behavior of equations that arise naturally in physics and engineering.