Problem 41 - Calculus Early Transcendentals

Question

Find the indicated derivative in two ways: a. Replace \(x\) and \(y\) to write \(z\) as a function of \(t\) and differentiate. b. Use the Chain Rule. \(z^{\prime}(t),\) where \(z=\frac{1}{x}+\frac{1}{y}, x=t^{2}+2 t,\) and \(y=t^{3}-2\)

Step-by-Step Solution

Answer

In this exercise, we were asked to find the derivative of the given function \(z\) with respect to \(t\) using two different methods: direct differentiation and the chain rule. We successfully found the derivative using both methods, which resulted in the same expression: $$\frac{dz}{dt} = -\frac{2t+2}{(t^2+2t)^2} - \frac{3t^2}{(t^3-2)^2}$$ This confirms that both methods are valid, and the derivative of the function \(z\) with respect to \(t\) is indeed the given expression.

Step 1: Substitute x and y with given functions of t

We are given the function \(z\) as \(z=\frac{1}{x}+\frac{1}{y}\), with \(x = t^2+2t\), and \(y = t^3-2\). By replacing \(x\) and \(y\) we obtain: $$z(t) = \frac{1}{t^2+2t}+\frac{1}{t^3-2}$$ Now, we can compute the derivative of \(z\) with respect to \(t\) using the first method.

Step 2: Differentiate z(t) with respect to t using the first method

We differentiate \(z(t)\) with respect to \(t\): $$\frac{dz}{dt} = \frac{d}{dt}\left(\frac{1}{t^2+2t}\right) + \frac{d}{dt}\left(\frac{1}{t^3-2}\right)$$ Using the power rule and the chain rule for derivatives, we find: $$\frac{dz}{dt} = -\frac{2t+2}{(t^2+2t)^2} - \frac{3t^2}{(t^3-2)^2}$$ This is the derivative of \(z\) with respect to \(t\) found by the first method. Now, let's find the derivative using the chain rule and compare the results to ensure they are identical.

Step 3: Apply the Chain Rule to differentiate z with respect to t

First, observe that: $$\frac{\partial z}{\partial x}=-\frac{1}{x^2}$$ $$\frac{\partial z}{\partial y}=-\frac{1}{y^2}$$ Also, we have: $$\frac{dx}{dt}=2t+2$$ $$\frac{dy}{dt}=3t^2$$ Now, applying the chain rule: $$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$$ $$\frac{dz}{dt} = -\frac{1}{x^2}(2t+2) - \frac{1}{y^2}(3t^2)$$

Step 4: Substitute x and y back with given functions of t

Now, we substitute \(x = t^2+2t\) and \(y = t^3-2\) back into the equation: $$\frac{dz}{dt} = -\frac{1}{(t^2+2t)^2}(2t+2) - \frac{1}{(t^3-2)^2}(3t^2)$$ As we can see, this result is the same as the one we obtained in Step 2.

Step 5: Conclusion

We have found the derivative of the given function \(z\) with respect to \(t\) using two different methods: direct differentiation and the chain rule. Both methods yielded the same result: $$\frac{dz}{dt} = -\frac{2t+2}{(t^2+2t)^2} - \frac{3t^2}{(t^3-2)^2}$$

Key Concepts

Chain Rule

The Chain Rule is a fundamental principle in calculus used to find the derivative of composite functions.
A composite function occurs when one function is nested inside another, such as in the situation with the equation for \(z\).
For example, when calculating the derivative of \(z\), which is a function of \(x\) and \(y\), where \(x\) and \(y\) themselves are functions of \(t\), the Chain Rule comes into play.
Instead of differentiating the entire composite function at once, the Chain Rule allows us to break it down. This involves calculating the partial derivatives of \(z\) with respect to \(x\) and \(y\), then multiplying these by the derivatives of \(x\) and \(y\) with respect to \(t\), respectively.
Hence, the process is as follows:

• Find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\)
• Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
• Combine these using \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\)

This technique is not only efficient but simplifies complex differentiation problems by tackling them in manageable parts.

Partial Derivative

Partial derivatives are used to assess the rate of change of a function with respect to one of its variables while keeping the others constant.
In contexts where functions depend on more than one variable, partial derivatives allow for the isolation of effects pertaining to a single variable.
For the given exercise, \(z\) is initially expressed in terms of \(x\) and \(y\). The partial derivative of \(z\) with respect to \(x\) is used to understand how \(z\) alters as \(x\) changes, assuming \(y\) remains constant, given by \(\frac{\partial z}{\partial x} = -\frac{1}{x^2}\). Similarly, for \(y\), it's \(\frac{\partial z}{\partial y} = -\frac{1}{y^2}\).
These derivatives play a crucial part in applying the Chain Rule because they provide the basis for how the overall rate of change will incorporate changes in each variable:

• Focus is on one variable at a time
• Useful for functions of multiple variables

Substitution Method

Substitution is a handy method to transform a function into a more manageable form for differentiation or integration.
In this exercise, substitution was used initially for both \(x\) and \(y\) to express \(z\) directly as a function of \(t\).
By substituting \(x = t^2 + 2t\) and \(y = t^3 - 2\) into the expression for \(z\), we obtained a new function of \(t\): \[ z(t) = \frac{1}{t^2 + 2t} + \frac{1}{t^3 - 2} \].
This approach simplifies the differentiation process because working with a single variable, \(t\), is often easier than dealing with the intersection of multiple variables. The key steps to keep in mind:

• Identify the components for substitution
• Rewrite the original function in one variable
• Differentiate the simplified function

This results in a straightforward calculation for the derivative of \(z\) with respect to \(t\), capturing all necessary interactions of \(x\) and \(y\) inherently.

Function Differentiation

Differentiation is a core process in calculus aimed at determining the rate at which one variable changes in relation to another, called the derivative.
For the function \(z\), which is expressed through algebraic fractions in terms of \(x\) and \(y\), finding the derivative with respect to \(t\) involves determining how \(z\) evolves as \(t\) changes.
In the exercise, the differentiation process was carried out in two main ways:

• Direct differentiation after expressing \(z\) as a function of \(t\)
• Applying the Chain Rule for differentiating the composite function

The calculation of derivatives like \(\frac{dz}{dt}\) provides critical insight into the behavior of \(z\) as the independent variable \(t\) shifts. In physics and engineering, such operations help in predicting future states or the overall dynamics of a system. Let's not forget that:

• Derivatives represent a slope of a curve at a point
• Essential for understanding changes and trends in functions