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Chapter 13: Problem 4

Use an appropriate form of the chain rule to find \(d z / d t\). $$ z=\sqrt{1+x-2 x y^{4}} ; x=\ln t, y=t $$

Short Answer

Expert verified
\( \frac{dz}{dt} = \frac{1}{2\sqrt{1 + \ln t - 2\ln t \cdot t^4}} \left(\frac{1}{t} - 8t^3 \ln t - 2t^3 \right) \).

Step by step solution

01

Identify the Composite Function

Notice that the given function \( z = \sqrt{1+x-2xy^4} \) is a composite function because the inner function is \( u = 1 + x - 2xy^4 \), and the outer function is \( z = \sqrt{u} \). This indicates the use of the Chain Rule in differentiation.
02

Differentiate the Outer Function

Differentiate the function \( z = \sqrt{u} \) with respect to \( u \) using the derivative of a square root. The derivative is: \( \frac{dz}{du} = \frac{1}{2\sqrt{u}} \).
03

Differentiate the Inner Function with Respect to \( t \)

The inner function is \( u = 1 + x - 2xy^4 \). Substitute \( x = \ln t \) and \( y = t \).Therefore, \( u = 1 + \ln t - 2\ln t \cdot t^4 \). Differentiating \( u \) with respect to \( t \) gives: \( \frac{du}{dt} = \frac{d}{dt}(1 + \ln t - 2\ln t \cdot t^4) = \frac{1}{t} - 2(\ln t \cdot 4t^3) - 2t^4 \cdot \frac{1}{t} \). This simplifies to: \( \frac{du}{dt} = \frac{1}{t} - 8t^3 \ln t - 2t^3 \).
04

Apply the Chain Rule to Find \( \frac{dz}{dt} \)

According to the chain rule, \( \frac{dz}{dt} = \frac{dz}{du} \cdot \frac{du}{dt} \).Using the results from Steps 2 and 3: \( \frac{dz}{dt} = \frac{1}{2\sqrt{1 + \ln t - 2\ln t \cdot t^4}} \cdot \left(\frac{1}{t} - 8t^3 \ln t - 2t^3 \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Function
Composite functions are functions composed of two or more simpler functions. In the given exercise, the function for **z** is represented as a combination of two layers. It's crucial to identify each layer when working on differentiation problems involving the Chain Rule. Here's a breakdown:
  • The inner function is what you evaluate first. In this case, it is \( u = 1 + x - 2xy^4 \), where **x** and **y** have their own separate definitions.
  • The outer function is then applied to the result of the first function. Here, **z** is the outer function described as \( z = \sqrt{u} \).
This layered structure is what makes the function "composite," as we have a "function of a function." Recognizing a composite function allows you to appropriately use the Chain Rule for differentiation.
Differentiation
Differentiation is the mathematical process of finding the rate at which a function is changing at any given point. In this problem, we're looking to differentiate **z** with respect to **t**. This is not straightforward due to the complex relationship between **z**, **x**, and **y**.For differentiation:
  • Differentiate the outer function \( z = \sqrt{u} \) with respect to \( u \). By applying the derivative of the square root, we get \( \frac{dz}{du} = \frac{1}{2\sqrt{u}} \).
  • Next, differentiate the inner function \( u = 1 + \ln t - 2 \ln t \cdot t^4 \) with respect to \( t \). This requires using both the product rule and the natural logarithm rule.
We apply these differentiation steps carefully, as each function's complexity can lead to small mistakes. The Chain Rule ultimately ties these derivatives together.
Calculus
Calculus is a fundamental area of mathematics focused on change and motion, dealing with concepts such as derivatives and integrals. Within calculus, the Chain Rule is essential for handling composite functions.In this context:
  • The Chain Rule states that if a variable **z** depends on a variable **u**, which in turn depends on **t**, then the derivative of **z** with respect to **t** can be found by multiplying the derivative of **z** with respect to **u** by the derivative of **u** with respect to **t**.
  • Symbolically, this is represented as \( \frac{dz}{dt} = \frac{dz}{du} \cdot \frac{du}{dt} \).
By applying the Chain Rule, we bring together the differentiated pieces of our composite function. Calculus, thus, provides a toolkit for solving problems involving changing rates and complex functional relationships.

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Most popular questions from this chapter

Find an equation for the tangent plane and parametric equations for the normal line to the surface at the point \(P\). $$ x^{2} y-4 z^{2}=-7 ; \quad P(-3,1,-2) $$

Find the directional derivative of \(f\) at \(P\) in the direction of a vector making the counterclockwise angle \(\theta\) with the positive \(x\) -axis. $$ f(x, y)=\frac{x-y}{x+y} ; P(-1,-2) ; \theta=\pi / 2 $$

Find the absolute extrema of the given function on the indicated closed and bounded set \(R .\) $$ f(x, y)=x^{2}+2 y^{2}-x ; R \text { is the disk } x^{2}+y^{2} \leq 4 $$

Determine whether the statement is true or false. Explain your answer. If the tangent plane to the level surface of \(F(x, y, z)\) at the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) is also tangent to a level surface of \(G(x, y, z)\) at \(P_{0}\), then \(\nabla F\left(x_{0}, y_{0}, z_{0}\right)=\nabla G\left(x_{0}, y_{0}, z_{0}\right)\)

A particle moves along a path \(C\) given by the equations \(x=t\) and \(y=-t^{2}\). If \(z=x^{2}+y^{2}\), find \(d z / d s\) along \(C\) at the instant when the particle is at the point \((2,-4)\).
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