Question

Find \(d y / d x\) by implicit differentiation. \(x^{2}+y^{2}=9\)

Short answer

The derivative of \(y\) with respect to \(x\) is \(dy/dx=-x/y\).

Step-by-step solution

Differentiation

Begin by differentiating each side of the equation \(x^{2}+y^{2}=9\). Derivative of \(x^2\) with respect to x is \(2x\). Derivative of \(y^2\) with respect to x is \(2y(dy/dx)\) by the chain rule. The right hand side derivative of a constant is 0.

Apply the Chain Rule

So, we have the equation: \(2x+2y(dy/dx)=0\). This equation is derived from differentiating each side of the equation (according to the chain rule) and setting it equal to 0.

Solve for dy/dx

Finally, solve for \(dy/dx\). We want to isolate \(dy/dx\) on one side. To do this, first subtract \(2x\) from both sides of the equation to get: \(2y(dy/dx)=-2x\). Then divide by \(2y\) on both sides to get: \(dy/dx = -\frac{x}{y}\).

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus, used for computing the derivative of a composite function. The basic idea of the chain rule is to break down a complicated function into simpler parts, differentiate each part, and then combine the derivatives to obtain the overall derivative. When looking at functions where one variable is expressed in terms of another, as with implicit functions like the equation of a circle, \(x^{2}+y^{2}=9\), the chain rule becomes essential.

When implicitly differentiating \(y^{2}\) with respect to \(x\), you consider \(y\) as a function of \(x\) (\(y(x)\)). You then apply the chain rule, which in this case turns out to be \(2y\frac{dy}{dx}\). This part of the derivative reflects the fact that as \(x\) changes, \(y\) is not constant but changes in a way that's dependent on \(x\). Hence, the derivative of \(y^{2}\) isn't just \(2y\) but \(2y\frac{dy}{dx}\) to account for the change in \(y\) as \(x\) varies.

Derivative of a Constant

In calculus, it's important to remember that the derivative of a constant is always zero. This concept is fundamentally rooted in the definition of a derivative, which is a measure of how a function changes as its input changes. Since a constant does not change, its rate of change is zero.

In the equation given, \(x^{2}+y^{2}=9\), the number 9 is a constant and does not involve the variable \(x\). Hence, when differentiating each side of the equation with respect to \(x\), the derivative of the right-hand side, 9, is zero. This is reflected in the initial steps of solving the exercise, indicating the importance of recognizing constants when solving for derivatives.

Solving for Derivatives

Solving for derivatives, especially in the context of implicit differentiation, requires algebraic manipulation to isolate the derivative term, \(\frac{dy}{dx}\). After applying the chain rule to differentiate the equation \(x^{2}+y^{2}=9\), we get \(2x+2y\frac{dy}{dx}=0\). To find the expression for \(\frac{dy}{dx}\), you have to rearrange the equation and isolate \(\frac{dy}{dx}\) on one side. This involves separation and solving of algebraic terms.

First, you subtract \(2x\) from both sides, yielding \(2y\frac{dy}{dx}=-2x\). Then, you divide both sides by \(2y\) to find that \(\frac{dy}{dx}=-\frac{x}{y}\). Mastery of this process is crucial, as it allows you to solve for \(\frac{dy}{dx}\), which represents the rate at which \(y\) changes with respect to \(x\) in the implicitly defined function.

Differentiation of Power Functions

Differentiation of power functions involves applying basic rules of derivatives to functions of the form \(x^n\), where \(n\) is any real number. The derivative of such a power function follows the rule: if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). This is sometimes referred to as the power rule.

In the context of our exercise, \(x^{2}\) is a power function with \(n=2\). When differentiating \(x^{2}\) with respect to \(x\), we apply the power rule to get the derivative, which is \(2x^{(2-1)}\) or simply \(2x\). This rule greatly simplifies the process of finding derivatives and is essential for students to learn and apply when working with power functions.