/*! 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 9 Use the chain rule to differenti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 9

Use the chain rule to differentiate (a) \(y=e^{x^{3}}\) (b) \(y=\ln \left(x^{4}+3 x^{2}\right)\)

Short Answer

Expert verified
a) \(3x^{2}e^{x^{3}}\), b) \(\frac{4x^3 + 6x}{x^4 + 3x^2}\)

Step by step solution

01

Identify the Outer Function

For the function given, identify the outermost function that is being applied. (a) The outer function here is the exponential function \(e^{u}\) where \(u=x^3\). (b) The outer function here is the natural logarithm \(\text{ln}(u)\) where \(u=x^4 + 3x^2\).
02

Identify the Inner Function

Determine the inner function that is being fed into the outer function. (a) The inner function \(u\) is \(x^3\). (b) The inner function \(u\) is \(x^4 + 3x^2\).
03

Differentiate the Outer Function

Differentiate the outer function with respect to the inner function \(u\). (a) For \(y = e^{u}\), \(\frac{dy}{du} = e^{u}\). (b) For \(y = \text{ln}(u)\), \(\frac{dy}{du} = \frac{1}{u}\).
04

Differentiate the Inner Function

Differentiate the inner function with respect to \(x\). (a) For \(u = x^3\), \(\frac{du}{dx} = 3x^{2}\). (b) For \(u = x^4 + 3x^2\), \(\frac{du}{dx} = 4x^{3} + 6x\).
05

Apply the Chain Rule

Combine the derivatives from Steps 3 and 4 using the chain rule \((\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx})\). (a) \(\frac{dy}{dx} = e^{x^3} \times 3x^{2} = 3x^{2}e^{x^{3}}\). (b) \(\frac{dy}{dx} = \frac{1}{x^4 + 3x^2} \times (4x^{3} + 6x) = \frac{4x^3 + 6x}{x^4 + 3x^2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Outer Function
The outer function is the function that is applied last when evaluating your given function. Think of it as wrapping another function around an existing one.
For example, in the case of the function \(y = e^{x^3}\), the outermost function is the exponential function, \(e^u\), where \(u = x^3\).
Meanwhile, for \(y = \text{ln}(x^4 + 3x^2)\), the outer function is the natural logarithm, \(\text{ln}(u)\), with \(u = x^4 + 3x^2\). Identifying the outer function correctly is crucial because its derivative will dictate part of the final expression for the solution.
Inner Function
The inner function is the one that goes into the outer function. In essence, it鈥檚 what you evaluate before applying the outer function.
For instance, in the function \(y = e^{x^3}\), the inner function is \(x^3\).
In the function \(y = \text{ln}(x^4 + 3x^2)\), the inner function is \(x^4 + 3x^2\). Identifying the inner function is fundamental, as you鈥檒l need to differentiate it separately from the outer function.
Derivative Calculation
After identifying the outer and inner functions, the next step is to calculate their derivatives individually.
When differentiating the outer function, you take its derivative with respect to the inner function.

For example:
  • For \(e^{u}\), \(\frac{dy}{du} = e^{u}\).
  • For \(\text{ln}(u)\), \(\frac{dy}{du} = \frac{1}{u}\).
Next, differentiate the inner function with respect to \(- x\).

In our example:
  • Given \(- u = x^3\), then \(\frac{du}{dx} = 3 x^2\).
  • For \(- u = x^4 + 3x^2\), \(\frac{du}{dx} = 4x^3 + 6x\).
Don't mix up the order! Correct differentiation ensures accurate results.
Step-by-Step Solution
Applying what you have learned through differentiation of the outer and inner functions, we now use the chain rule to get the final derivative.
The chain rule states \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\). You simply multiply the derivatives obtained in the previous steps.

For our problems:
  • \((y = e^{x^3})\) becomes \(\frac{dy}{dx} = e^{x^3} \times 3x^2 = 3x^2e^{x^{3}}\).
  • Meanwhile, \((y = \text{ln}(x^4 + 3x^2))\) turns to \(\frac{dy}{dx} = \frac{1}{x^4 + 3x^2} \times (4x^3 + 6x) = \frac{4x^3 + 6x}{x^4 + 3x^2}\).
By following these steps, you can confidently apply the chain rule to differentiate complex functions!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Differentiate (a) \(y=x(x-3)^{4}\) (b) \(y=x \sqrt{(2 x-3)}\) (c) \(y=\frac{x}{x+5}\) (d) \(y=\frac{x}{x^{2}+1}\)

For each of the graphs (a) \(y=\sqrt{x}\) (b) \(y=x \sqrt{x}\) (c) \(y=\frac{1}{\sqrt{x}}\) \(A\) is the point where \(x=4\), and \(B\) is the point where \(x=4.1 .\) In each case find (i) the \(y\) coordinates of \(A\) and \(B\). (ii) the gradient of the chord \(A B\) (iii) the value of \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(A\). Compare your answers to parts (ii) and (iii).

(a) Attempt to use Maple to plot a graph of the function \(y=1 / x\) over the range \(-4 \leq x \leq 4\). What difficulty do you encounter? Explain briefly why this has occurred for this particular function. (b) One way of avoiding the difficulty in part (a) is to restrict the range of the \(y\) values. Produce a plot by typing $$ \operatorname{plot}(1 / x, x=-4 \ldots 4, y=-3 \ldots 3) $$ (c) Use the approach suggested in part (b) to plot a graph of the curve $$ y=\frac{x-3}{(x+1)(x-2)} $$ on the interval \(-2 \leq x \leq 6\). Use calculus to find all of the stationary points.

Differentiate (a) \(y=5 x^{2}\) (b) \(y=\frac{3}{x}\) (c) \(y=2 x+3\) (d) \(y=x^{2}+x+1\) (e) \(y=x^{2}-3 x+2\) (f) \(y=3 x-\frac{7}{x}\) (g) \(y=2 x^{3}-6 x^{2}+49 x-54\) (h) \(y=a x+b\) (i) \(y=a x^{2}+b x+c\) (j) \(y=4 \sqrt{x}-\frac{3}{x}+\frac{7}{x^{2}}\)

A firm's short-run production function is given by $$ Q=300 L^{2}-L^{4} $$ where \(L\) denotes the number of workers. Find the size of the workforce that maximizes the average product of labour and verify that at this value of \(L\) $$ \mathrm{MP}_{L}=\mathrm{AP}_{L} $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.