Question

Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the given point. \(y^{3}-x^{2}=4, \quad(2,2)\)

Short answer

The derivative of the function at the point (2,2) is 1/3.

Step-by-step solution

Differentiate both sides of the equation

First, take the derivative of both sides of the equation \(y^{3} - x^{2} = 4\) with respect to \(x\). This gives \(\frac{d}{dx}(y^{3} - x^{2}) = \frac{d}{dx}(4)\).

Use chain rule for y and power rule for x

Now apply the chain rule to differentiate \(y^{3}\), and the power rule to differentiate \(-x^{2}\). When differentiating \(y^{3}\), the chain rule is used because although we're differentiating with respect to \(x\), \(y\) is a function of \(x\). The chain rule gives \(3y^{2} \cdot \frac{dy}{dx}\). The power rule gives \(-2x\). So we get \(3y^{2}\frac{dy}{dx} - 2x = 0\).

Solve the equation for dy/dx

Reorganize the above equation, we get \(\frac{dy}{dx} = \frac{2x}{3y^{2}}\).

Substitute the given point

Now evaluate \(\frac{dy}{dx}\) at the point (2,2). Substitute \(x = 2\) and \(y = 2\), giving \(\frac{dy}{dx} = \frac{2\cdot2}{3\cdot2^{2}}\).

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus, particularly useful in differentiating composite functions. In the context of implicit differentiation, where a function is indirectly defined, the chain rule helps in taking derivatives with respect to one variable when other variables are involved.
The chain rule states that when differentiating a function that is nested within another, you take the derivative of the outer function and multiply it by the derivative of the inner function. In our exercise given by the equation \(y^3 - x^2 = 4\), \(y\) is considered a function of \(x\).
To differentiate \(y^3\), we apply the chain rule by:

• First differentiating the outer function \(3y^2\), treating \(y\) as a constant temporarily.
• Then multiplying by \(\frac{dy}{dx}\), which is the derivative of \(y\) with respect to \(x\).

This yields \(3y^2 \cdot \frac{dy}{dx}\). By performing these steps diligently, we can differentiate complex expressions that involve nested relationships between variables.

Power Rule

The power rule is one of the simplest derivative rules, yet almost universally applicable in calculus. It states that for any function \(x^n\), the derivative is \(nx^{n-1}\). This rule is used whenever we have a power of \(x\) that needs differentiating.
In our initial equation \(y^3 - x^2 = 4\), we apply the power rule to the \(-x^2\) term. Differentiating \(-x^2\) with respect to \(x\), we get:

• Bring down the exponent: multiply the coefficient by the exponent, which is 2 here, to get \(-2x\).
• Decrease the exponent by one: reducing \(x^2\) by 1 means it becomes \(x^{2-1} = x^1\).

Thus, the derivative of \(-x^2\) is \(-2x\). This step is crucial as it allows us to translate differentials into more manageable algebraic expressions. Alongside the chain rule for \(y^3\), it enables the equation to be rearranged and solved for \(\frac{dy}{dx}\).

Derivative Evaluation

Evaluating a derivative involves simplifying and substituting known values to find the rate of change at a specific point. From our equation, after implicit differentiation, we obtained \(\frac{dy}{dx} = \frac{2x}{3y^2}\). The next step is to evaluate this derivative at the given point \((2, 2)\).
To do this:

• Substitute \(x = 2\) and \(y = 2\) into the derivative \(\frac{dy}{dx} = \frac{2x}{3y^2}\).
• Perform the needed arithmetic: \(\frac{2 \times 2}{3 \times 2^2} = \frac{4}{12}\), which simplifies to \(\frac{1}{3}\).

This result, \(\frac{1}{3}\), represents the slope of the tangent line to the curve defined by the equation \(y^3 - x^2 = 4\) at the specific point. Performing these evaluations verifies whether our implicit differentiation and subsequent simplifications were correctly executed, providing valuable insights into the behavior of the function at particular coordinates.