/*! 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 62 Differentiate with respect to th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate with respect to the independent variable. $$ f(x)=\frac{\sqrt{5 x\left(1+x^{2}\right)}}{\sqrt{2}} $$

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{15x^2 + 5}{2\sqrt{10x^3 + 10x}}. \)

Step by step solution

01

Setup the Function for Differentiation

The function given is \( f(x) = \frac{\sqrt{5x(1+x^2)}}{\sqrt{2}} \). This can be rewritten as \( f(x) = \frac{1}{\sqrt{2}} \cdot \sqrt{5x + 5x^3} \). For differentiation, we need to use the chain rule and, potentially, the product rule.
02

Differentiate Using the Chain Rule

The function can be modeled as \( u(x) = \sqrt{5x^3 + 5x} \) where \( f(x) = \frac{1}{\sqrt{2}} u(x) \). The derivative \( u'(x) \) uses the chain rule: \( \frac{1}{2\sqrt{5x^3 + 5x}} \cdot (15x^2 + 5) \).
03

Apply the Constant Multiple Rule

As \( f(x) = \frac{1}{\sqrt{2}} u(x) \), the derivative \( f'(x) \) will thus be \( \frac{1}{\sqrt{2}} \cdot u'(x) \). Substitute \( u'(x) \) from Step 2 to get the derivative: \[ f'(x) = \frac{1}{\sqrt{2}} \cdot \frac{15x^2 + 5}{2\sqrt{5x^3 + 5x}} \].
04

Simplify the Expression

Combine and simplify the constants and expressions if possible. The final derivative becomes \[ f'(x) = \frac{15x^2 + 5}{2\sqrt{2} \cdot \sqrt{5x^3 + 5x}} = \frac{15x^2 + 5}{2\sqrt{10x^3 + 10x}}. \]

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 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
The chain rule is essential for differentiating complex functions. It's used when you have a function inside another function. In other words, when you want to differentiate the composition of two functions.
To apply the chain rule effectively, you must identify both the outer function and the inner function. The rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function itself.
  • If you have a function of the form \( g(f(x)) \), then the derivative is \( g'(f(x)) \cdot f'(x) \).
  • Always take the derivative of the outer function, leaving the inner function as it is, and then multiply by the derivative of the inner function.
In the original exercise, when differentiating \( u(x) = \sqrt{5x^3 + 5x} \), the inner function is \( 5x^3 + 5x \), and its derivative is \( 15x^2 + 5 \). You then multiply by the derivative of the inner part, which is captured using the chain rule.
Product Rule
The product rule is another differentiation technique useful in situations where two functions are multiplied together. In calculus, this rule simplifies finding the derivative of a product of two functions.
The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), their product \( u(x) \cdot v(x) \) has a derivative:
  • \( (uv)' = u'v + uv' \)
This means you take the derivative of the first function multiplied by the second function as it is, plus the first function as it is multiplied by the derivative of the second function. In the exercise, while you might expect needing the product rule, since the main operation is involving a constant and a function, only the constant multiple rule was necessary. However, understanding the product rule can clarify when to apply it in more complex cases.
Differentiation Techniques
Differentiation techniques are essential for handling various types of functions you face in calculus. These techniques include the chain rule, product rule, quotient rule, and others, each catering to specific functional patterns.
Here's a quick overview:
  • Chain Rule: Useful for nested functions, where you differentiate the outer function and multiply by the derivative of the inner function.
  • Product Rule: Applicable when differentiating a product of two functions. Remember the formula: \((uv)' = u'v + uv'\).
  • Quotient Rule: Essential when differentiating a division of two functions. The formula is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
  • Constant Multiple Rule: If a function is multiplied by a constant, differentiate as usual and then multiply the result by that constant.
Applying the right technique makes calculating derivatives more efficient and less error-prone. In the exercise, different rules were combined to achieve the final result, showcasing the importance of mastering these techniques for solving calculus problems effectively.

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

Suppose that the per capita growth rate of a population is \(2 \% ;\) that is, if \(N(t)\) denotes the population size at time \(t\), then $$\frac{1}{N} \frac{d N}{d t}=0.02$$ Suppose also that the population size at time \(t=2\) is equal to 50. Use a linear approximation to compute the population size at time \(t=2.1\).

Find the equation of the tangent line to the curve \(y=\sqrt{x}\) at the point \((4,2)\).

Assume that the measurement of \(x\) is accurate within \(2 \% .\) In each case, determine the error \(\Delta f\) in the calculation of \(f\) and find the percentage error \(100 \frac{\Delta f}{f} .\) The quantities \(f(x)\) and the true value of \(x\) are given. $$ f(x)=\frac{1}{1+x}, x=4 $$

4Suppose a particle moves along a straight line. The position at time \(t\) is given by $$ s(t)=3 t-t^{2}, \quad t \geq 0 $$ where \(t\) is measured in seconds and \(s(t)\) is measured in meters. (a) Graph \(s(t)\) for \(t \geq 0\). (b) Use the graph in (a) to answer the following questions: (i) Where is the particle at time \(0 ?\) (ii) Is there another time at which the particle visits the location where it was at time \(0 ?\) (iii) How far to the right on the straight line does the particle travel? (iv) How far to the left on the straight line does the particle travel? (v) Where is the velocity positive? where negative? equal to 0 ? (c) Find the velocity of the particle. (d) When is the velocity of the particle equal to \(1 \mathrm{~m} / \mathrm{s}\) ?

The following limit represents the derivative of a function \(f\) at the point \((a, f(a))\) : $$ \lim _{h \rightarrow 0} \frac{4(a+h)^{3}-4 a^{3}}{h} $$

See all solutions

Recommended explanations on Biology Textbooks

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.