Problem 241 - Calculus Volume 3

Question

Let \(z=e^{1-x y}, x=t^{1 / 3},\) and \(y=t^{3} .\) Find \(\frac{d z}{d t}\)

Step-by-Step Solution

Answer

\(\frac{dz}{dt} = -\frac{10}{3}t^{7/3}e^{1 - t^{10/3}}.\)

Step 1: Express z as a function of t

We know that \(z = e^{1 - xy}\), \(x = t^{1/3}\), and \(y = t^3\). Substitute the expressions for \(x\) and \(y\) in terms of \(t\) into the equation for \(z\):\[z = e^{1 - (t^{1/3})(t^3)} = e^{1 - t^{10/3}}.\]This expresses \(z\) purely as a function of \(t\).

Step 2: Differentiate z with respect to t

Now differentiate \(z = e^{1 - t^{10/3}}\) with respect to \(t\). We apply the chain rule. Let \(u = 1 - t^{10/3}\), so \(z = e^u\). Differentiate \(z\) with respect to \(u\):\[\frac{dz}{du} = e^u = e^{1 - t^{10/3}}.\]Then differentiate \(u\) with respect to \(t\):\[\frac{du}{dt} = -\frac{10}{3}t^{7/3}.\]Thus, the derivative of \(z\) with respect to \(t\) is given by the product:\[\frac{dz}{dt} = \frac{dz}{du} \cdot \frac{du}{dt} = e^{1 - t^{10/3}} \cdot \left(-\frac{10}{3}t^{7/3}\right).\]

Step 3: Simplify the resulting expression

Simplify the expression for \(\frac{dz}{dt}\). We have:\[\frac{dz}{dt} = -\frac{10}{3}t^{7/3}e^{1 - t^{10/3}}.\]There are no further simplifications possible, so this is our final expression for the derivative.

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental technique in calculus used for differentiating composite functions. A composite function is one where you have one function inside another. To understand this better, imagine a function that contains another function; for example, if we have a function \( f(g(x)) \), where \( g(x) \) is nestled inside \( f \). The idea is to differentiate the outer function and then multiply it by the derivative of the inner function.

When applying the chain rule, you can break it down into steps:

• Identify the inner function and the outer function.
• Take the derivative of the outer function with respect to the inner function.
• Multiply the result by the derivative of the inner function with respect to \( x \).

This is exactly what we do in the solution by defining \( u = 1 - t^{10/3} \) as the inner function and \( z = e^u \) as the outer function. This approach helps manage the complexity of finding derivatives of more elaborate expressions.

Exponential Functions

Exponential functions are expressions where a constant base is raised to a variable exponent. The most common base for these functions is Euler's number, \( e \), approximately equal to 2.718. This function, \( e^x \), is unique because it is its own derivative.

In the context of the given exercise, exponential functions characterize how changes in input (\( t \) in this case) can dramatically affect the output. The exercise uses \( z = e^{1 - t^{10/3}} \), showing the sensitivity of \( z \) to changes in \( t \).

To differentiate these functions, we adhere to their unique properties. Particularly, if \( y = e^x \), then the derivative \( \frac{dy}{dx} = e^x \) remains unchanged. This property simplifies many calculus problems, as you can perform differentiation with fewer transformations.

Steps in Derivative Computation

Calculating derivatives is a central process in calculus, allowing us to determine the rate at which one quantity changes with respect to another. The exercise at hand focuses on finding \( \frac{dz}{dt} \) for \( z = e^{1 - t^{10/3}} \).

Here's a breakdown of the derivative computation steps:

• Recognize that \( z \) includes an exponential function, requiring us to track changes in the exponent, \( u \).
• Substitute \( u = 1 - t^{10/3} \). Compute the derivative of \( z \) with respect to \( u \) using the property of exponential functions, resulting in \( \frac{dz}{du} = e^u \).
• Next, differentiate \( u \) with respect to \( t \), yielding \( \frac{du}{dt} = -\frac{10}{3}t^{7/3} \).
• Finally, apply the chain rule to find \( \frac{dz}{dt} \) by multiplying \( \frac{dz}{du} \) and \( \frac{du}{dt} \).

The final result, \( -\frac{10}{3}t^{7/3}e^{1 - t^{10/3}} \), gracefully combines these elements, culminating in the precise rate at which \( z \) changes as \( t \) shifts.