/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Find \(\partial z / \partial u\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(\partial z / \partial u\) and \(\partial z / \partial v\) when \(u=1, v=-2\) if \(z=\ln q\) and \(q=\sqrt{v+3} \tan ^{-1} u.\)

Short Answer

Expert verified
\( \partial z / \partial u = \frac{2}{\pi}; \partial z / \partial v = 1 \) at \( u = 1, v = -2 \).

Step by step solution

01

Understand the problem

We need to find the partial derivatives \( \partial z / \partial u \) and \( \partial z / \partial v \) at the specific values \( u = 1 \) and \( v = -2 \), given the function \( z = \ln q \) where \( q = \sqrt{v+3} \tan^{-1} u \).
02

Find the expression for z in terms of u and v

Given \( q = \sqrt{v+3} \tan^{-1} u \), we substitute this into the equation for \( z \): \( z = \ln(q) = \ln(\sqrt{v+3} \tan^{-1} u) \).
03

Differentiate z with respect to u

Using the chain rule, \( \partial z / \partial u = \frac{1}{q} \cdot \frac{\partial q}{\partial u} \). First, find \( \partial q / \partial u = \sqrt{v+3} \cdot \frac{1}{1+u^2} \). Thus, \( \partial z / \partial u = \frac{\sqrt{v+3}}{(\sqrt{v+3} \tan^{-1} u)(1+u^2)} = \frac{1}{\tan^{-1} u (1+u^2)}\).
04

Differentiate z with respect to v

Again, use the chain rule: \( \partial z / \partial v = \frac{1}{q} \cdot \frac{\partial q}{\partial v} \). Firstly, \( \partial q / \partial v = \frac{1}{2\sqrt{v+3}} \tan^{-1} u \). Thus, \( \partial z / \partial v = \frac{\tan^{-1} u}{2 (\sqrt{v+3})(\sqrt{v+3} \tan^{-1} u)} = \frac{1}{2(v+3)}\).
05

Substitute u = 1 and v = -2 into the partial derivatives

Substitute \( u = 1 \) and \( v = -2 \) into the expressions derived: \( \partial z / \partial u |_{(1,-2)} = \frac{1}{\tan^{-1} 1 (1+1^2)} = \frac{1}{\frac{\pi}{4} \cdot 2} = \frac{2}{\pi} \). And \( \partial z / \partial v |_{(1,-2)} = \frac{1}{2(-2+3)} = 1\).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus that allows us to differentiate composite functions. In the context of partial derivatives, the chain rule helps us differentiate functions that depend on multiple variables. For this exercise, we have specifically used the chain rule to find how changes in variables \(u\) and \(v\) affect another variable \(z\), through the intermediary variable \(q\).
By leveraging the chain rule, we first express the derivative of \(z\) with respect to either \(u\) or \(v\) in terms of \(q\), and then find the derivative of \(q\) with the concerned variable. By multiplying these two derivatives, we obtain the partial derivative of \(z\). This approach simplifies the process of dealing with composite functions in multivariable calculus.
Calculus Problem Solving
Approaching calculus problem-solving effectively involves understanding the problem, planning the steps, and executing the solution systematically. In this exercise, we began by understanding that the task was to find specific partial derivatives of a composite function at particular values. Recognizing these complexities early can help predict what techniques to apply.
  • Firstly, we identified the chain-rule as applicable due to the composite nature of \(z = \ln(q)\).
  • Then, careful substitution and differentiation were performed, knowing that our final derivatives should be evaluated at given points \(u = 1\) and \(v = -2\).
This strategic approach helps ensure all requisite steps are followed accurately and efficiently, resulting in a correct solution.
Multivariable Calculus
Multivariable calculus involves functions with more than one variable, which introduces additional complexity into differentiation. For this reason, functions like \(z = \ln(q)\) in the exercise not only depend on one variable but two or even more.
In the exercise, \(q\) is influenced both by \(u\) and \(v\), illustrating how different inputs can affect the output differently. Understanding the relationships between inputs and outputs is key. Additionally, multivariable calculus often makes use of partial derivatives, allowing us to explore how functions change with respect to one variable while keeping others constant. This capacity is essential when analyzing how various factors can independently affect outcomes.
Differentiation Techniques
Differentiation techniques in calculus provide tools to find rates of change, crucial for functions with complex dependencies. Here, we applied different techniques such as:
  • **Chain Rule**: For differentiating composite functions.
  • **Implicit Differentiation**: Useful in scenarios where functions are not straightforwardly expressed.
Notably in this problem, the chain rule played a central role. By differentiating \(q\) with respect to \(u\) and \(v\), we obtained expressions that allowed the calculation of \(\partial z/\partial u\) and \(\partial z/\partial v\). Knowing multiple differentiation techniques and understanding when to apply each is vital for tackling diverse mathematical problems efficiently.

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Most popular questions from this chapter

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