Question

Find \(d^{2} y / d x^{2}\) implicitly in terms of \(x\) and \(y\). \(x^{2}+y^{2}=4\)

Short answer

The second derivative \(d^{2}y/dx^{2}\) in terms of \(x\) and \(y\) is \(d^{2}y/dx^{2} = \frac{y + x^{2}}{y^{2}}\).

Step-by-step solution

Differentiate the Given Equation

For the equation \(x^{2}+y^{2}=4\), we will apply the derivative to each term. The derivative of \(x^{2}\) with respect to \(x\) is \(2x\). For \(y^{2}\) we use the chain rule, which means we first take the derivative of \(y^{2}\) with respect to \(y\) which gives us \(2y\) and then multiply by \(dy/dx\). That gives us \(2x + 2y \cdot \frac{dy}{dx} = 0.\)

Solve for dy/dx

Next express \(dy/dx\) in terms of \(x\) and \(y\) by solving above equation. (This is the first derivative). Solving for \(\frac{dy}{dx}\), we get \(\frac{dy}{dx} = -\frac{x}{y}\).

Differentiate once more to find the second derivative

Now we need to find the second derivative \(d^{2}y/dx^{2}\). We differentiate \(\frac{dy}{dx} = -\frac{x}{y}\) with respect to \(x\). Applying the quotient rule, \(d^{2}y/dx^{2}\) = \(\frac{y \cdot (1) - x \cdot dy/dx}{y^{2}}\) = \(\frac{y + x \cdot \frac{x}{y}}{y^{2}}\) = \(\frac{y + x^{2}}{y^{2}}\) considering that \(\frac{dy}{dx} = -\frac{x}{y}.\)

Key concepts

Second Derivative

When working with the concept of a second derivative in calculus, you're diving into an analysis of how the rate of change of a quantity is itself changing. Simply put, it's the derivative of the derivative, representing the acceleration of a function's rate of change.

To find the second derivative, as in our implicit differentiation problem with the equation x^{2}+y^{2}=4, you first differentiate to find the first derivative (dy/dx). Then, you differentiate this result once more with respect to x. This gives you the second derivative (d^{2}y/dx^{2}) which provides insight into the curvature or concavity of the function's graph. Quick acceleration changes suggest a sharp turn on the graph, indicative of a function's inflection points or extreme values.

Chain Rule

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. When a function y depends on another function u, which itself is a function of x (i.e., y(u(x))), then the chain rule lets you differentiate y with respect to x efficiently.

In the context of our problem, when we differentiate y^{2}, we acknowledge that y is a function of x — thus, we apply the chain rule. This involves taking the derivative of y^{2} with respect to y to get 2y and then multiplies this by dy/dx, symbolizing how y changes with x.

Quotient Rule

The quotient rule is another differentiation tool used when you're dealing with a function divided by another function. In mathematical terms, if you have a function in the form (u/v), then the derivative of this function with respect to x is (v(u') - u(v'))/(v^{2}), where u' and v' represent the derivatives of u and v, respectively.

Back to our exercise, in Step 3, when differentiating dy/dx = -x/y to find the second derivative, we applied the quotient rule. This step was crucial because dy/dx is essentially the ratio of two functions: -x (numerator) and y (denominator). Applying this rule provides the equation for the second derivative.

Derivative of a Function

The derivative of a function is the foundation of differential calculus. It quantifies the rate at which a function's value changes as its input changes. The derivative is represented by dy/dx for functions y(x) and indicates the slope of the function at any given point.

In our case, we're interested in implicit differentiation, which is used when functions are not defined explicitly as y=f(x). Our equation x^{2} + y^{2} = 4 expressed a relationship between x and y implicitly, with y not isolated on one side. By differentiating both sides with respect to x and solving for dy/dx, we find the derivative in terms of both x and y. This strategy is crucial in problems where explicit differentiation is not possible.