/*! 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 18 Find the derivative of the follo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivative of the following functions. See Example 4 of Section 3.1 for the derivative of \(\sqrt{x}\). $$f(s)=\frac{\sqrt{s}}{4}$$

Short Answer

Expert verified
Question: Find the derivative of the function $$f(s) = \frac{\sqrt{s}}{4}$$. Answer: The derivative of the given function is $$f'(s) = \frac{1}{8}s^{-\frac{1}{2}}$$.

Step by step solution

01

Calculate the derivative of \(\sqrt{s}\)

First, let's recall that the square root of any variable can be rewritten as an exponent. Thus, we can rewrite \(\sqrt{s}\) as \(s^{\frac{1}{2}}\). To find the derivative, we use the power rule for differentiation, which states: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\) Applying this rule, we get: $$\frac{d}{ds}(s^{\frac{1}{2}}) = \frac{1}{2}s^{\frac{1}{2}-1} = \frac{1}{2}s^{-\frac{1}{2}}$$
02

Apply the constant factor rule

Now, since our function is multiplied by a constant, we use the constant factor rule. This rule states that if you have a function \(f(x) = kg(x)\), where \(k\) is a constant, then the derivative \(f'(x) = kg'(x)\). In our case, the constant \(k = \frac{1}{4}\). Thus, we can rewrite the function as: $$f(s) = \frac{1}{4}s^{\frac{1}{2}}$$ Now applying the constant factor rule, we get the derivative of the function: $$f'(s) = \frac{1}{4} \times \frac{1}{2}s^{-\frac{1}{2}}$$
03

Simplify and find the final derivative

Finally, we simplify the derivative expression to get: $$f'(s) = \frac{1}{8}s^{-\frac{1}{2}}$$ This is the derivative of the given function.

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.

Power Rule
Differentiation is the process of finding a derivative, which tells us how a function changes at any point. The Power Rule is a fundamental differentiation technique used to find derivatives of polynomial expressions. Let's break it down in a simple way.
Imagine you have a function where the variable is raised to a power, like \(x^n\). The Power Rule states that the derivative of \(x^n\) is \(nx^{n-1}\). This means you bring the power down as a coefficient in front of the variable and then reduce the power by one.

Here's a quick example: if \(f(x) = x^3\), then applying the Power Rule gives us \(f'(x) = 3x^{2}\). You took 3 from the exponent, placed it in front, and reduced the original power by 1. This approach remains consistent no matter the value of \(n\), whether it's a whole number, fraction, or even a negative number.
The Power Rule is a powerful tool that makes differentiating polynomial functions straightforward and efficient.
Constant Factor Rule
Another crucial rule in differentiation is the Constant Factor Rule. This rule helps when our function is multiplied by a constant. It simplifies the process significantly since constants don’t change and can be moved outside of the differentiation process.
Let's consider a function like \(f(x) = kx^n\), where \(k\) is a constant. The Constant Factor Rule tells us that the derivative of this function is \(f'(x) = k \cdot nx^{n-1}\). Here, we simply apply the Power Rule to \(x^n\) and then multiply the resulting derivative by the constant \(k\).

For instance, if \(f(x) = 4x^2\), the derivative \(f'(x)\) would be \(4 \times 2x^{1}\) which simplifies to \(8x\). Since the constant doesn’t change when differentiating, it's often a straightforward multiplier for our final expression. This rule is essential for handling functions where constants are involved alongside variables.
Derivative of Square Root Function
The derivative of a square root function may initially seem daunting, but by converting it into a familiar form, the process becomes manageable. When we deal with a square root, such as \(\sqrt{x}\), we rewrite it as \(x^{1/2}\). This transformation allows us to employ the Power Rule for differentiation.
Once we have \(x^{1/2}\), applying the Power Rule gives the derivative as \(\frac{1}{2}x^{-1/2}\). This derivative can be interpreted as a scaling and reversing operator on the function. It shows how the output changes in response to changes in \(x\), specifically for functions shaped like square roots.

Thus, the Derivative of a Square Root Function is effectively about bringing the exponent down, reducing it, and handling the resultant negative exponent in terms of fractions if necessary. This systematic approach ensures that whatever the function’s form, we can dissect its behavior at any point quickly and accurately.

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

Calculate the derivative of the following functions (i) using the fact that \(b^{x}=e^{x \ln b}\) and (ii) by using logarithmic differentiation. Verify that both answers are the same. $$y=3^{x}$$

Find \(\frac{d y}{d x},\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\)

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no points of intersection (for \(x>0\) ), and the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 0 \) ). Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

Quotient Rule for the second derivative Assuming the first and second derivatives of \(f\) and \(g\) exist at \(x,\) find a formula for \(\frac{d^{2}}{d x^{2}}\left(\frac{f(x)}{g(x)}\right)\)

Suppose \(f\) is differentiable on an interval containing \(a\) and \(b\), and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x}\), show that \(c=\sqrt{a b}\), the geometric mean of \(a\) and \(b\), for \(a > 0\) and \(b > 0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a > 0\) and \(b > 0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

See all solutions

Recommended explanations on Math 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.