/*! 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 5 Identify the "inside function" \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Identify the "inside function" \(u=f(x)\) and the "outside function" \(y=g(u) .\) Then find \(d y / d x\) using the Chain Rule $$ y=\sqrt{e^{x}+\cos x} $$

Short Answer

Expert verified
The inside function is \(u = e^x + \cos x\), and the outside function is \(y = \sqrt u\). Using the Chain Rule, we find the derivative: $$ \frac{dy}{dx} = \frac{e^x - \sin x}{2 \sqrt{e^x + \cos x}} $$

Step by step solution

01

Identify the Inside and Outside Functions

First, we'll look at the given function: $$ y=\sqrt{e^{x}+\cos x} $$ The inside function is the one that is inside the square root. In this case, that function is: $$ u = e^x + \cos x $$ The outside function is the square root function which can be written as: $$ y = g(u) = \sqrt u $$
02

Find the Derivative of the Inside Function

Now we need to find the derivative of the inside function, i.e., \(\frac{du}{dx}\). $$ u = e^x + \cos x \\ \frac{du}{dx} = \frac{d}{dx} (e^x + \cos x) $$ Using basic rules of differentiation, we get: $$ \frac{du}{dx} = e^x - \sin x $$
03

Find the Derivative of the Outside Function

Next, we need to find the derivative of the outside function, i.e., \(\frac{dy}{du}\), where \(y = \sqrt u\). Rewriting the function to make the differentiation easier: $$ y = u^{\frac{1}{2}} \\ \frac{dy}{du} = \frac{1}{2} u^{-\frac{1}{2}} $$
04

Apply the Chain Rule

Now, we'll apply the Chain Rule to find \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$ Substitute the values we found in Steps 2 and 3: $$ \frac{dy}{dx} = \frac{1}{2} u^{-\frac{1}{2}} \cdot (e^x - \sin x) $$
05

Substitute the Inside Function back

Finally, substitute \(u\) back in terms of \(x\) using \(u = e^x + \cos x\): $$ \frac{dy}{dx} = \frac{1}{2} (e^x + \cos x)^{-\frac{1}{2}} \cdot (e^x - \sin x) $$ This is the derivative of the given function: $$ \frac{dy}{dx} = \frac{e^x - \sin x}{2 \sqrt{e^x + \cos x}} $$

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.

Inside Function
The 'inside function' is a key concept when applying the Chain Rule in calculus. When we deal with a composite function- a function within another function- the 'inside function' is the function that is inside another function. Think of it as the starting point of the nesting dolls, the smallest piece tucked inside all the others.

For instance, in the expression \( y=\sqrt{e^{x}+\cos x} \), the inside function is \( u = e^x + \cos x \), because it is within the square root. When differentiating a composite function, we first need to differentiate this inside function with respect to \( x \), which gives us its derivative. Understanding the role of the inside function is crucial for correctly applying the Chain Rule and finding the derivative of the overall composite function.
Outside Function

Peeling the Layers Off

The 'outside function' can be thought of as the outer layer of the nesting dolls. It's the function that has taken another function—our 'inside function'—as its input. With the composite function \( y=\sqrt{e^{x}+\cos x} \), the outside function is the square root operation, represented as \( y = \sqrt{u} \), where \( u \) embodies the inside function.

When we differentiate using the Chain Rule, we take the derivative of the outside function alone, pretending for a moment that the inside function is just any other variable. This part of the process is crucial because it sets the stage for combining the derivative of the outside function with that of the inside function to get the total derivative of the original composite function.
Derivative of a Function

The Heart of Change

The derivative of a function represents the rate at which the function is changing at any given point. It's the essential tool in calculus for understanding dynamic systems. In practical terms, if you visualize the graph of a function, the derivative at any point gives the slope of the tangent line to the graph at that point. This is the instantaneous rate of change of the function with respect to its variable.

The process of finding a derivative is known as differentiation, and there are various rules to do this, depending on the form of the function. For instance, the derivative of \( e^x \) is \( e^x \) itself, while the derivative of \( \cos x \) is \( -\sin x \). The ability to find derivatives precisely and efficiently is a cornerstone of solving many problems in mathematics and physics.
Rules of Differentiation

The Roadmap of Calculus

The rules of differentiation are the standardized methods used to find the derivative of a function. These rules provide a systematic approach for tackling various types of functions and their combinations. Here are a few cornerstone rules:
  • The Power Rule for derivatives states that the derivative of \( x^n \) is \( nx^{n-1} \).
  • The Product Rule tells us how to differentiate products of two functions.
  • The Quotient Rule is used to differentiate ratios of two functions.
  • And then, there's the Chain Rule, which is used for differentiating composite functions. It involves taking the derivative of the outside function multiplied by the derivative of the inside function.
Understanding these rules and when to apply each is vital for students as they represent the toolkit for problem-solving in differential calculus.

For our exercise, the Chain Rule is the hero, enabling us to differentiate a function nested within another function by systematically taking the derivative of each part, inside and out, and combining them to find the rate of change of the original composite function.

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

In Exercises, (a) find the equations of the tangent and the normal lines to the curve at the indicated point. (The normal line at a point on the curve is the line perpendicular to the tangent line at that point.) (b) Then use a graphing utility to plot the curve and the tangent and normal lines on the same screen. $$ x^{2}+y^{2}=9 ; \quad(-1,2 \sqrt{2}) $$

Potential of a Charged Disk The potential on the axis of a uniformly charged disk is $$ V(r)=\frac{\sigma}{2 \varepsilon_{0}}\left(\sqrt{r^{2}+R^{2}}-r\right) $$ where \(\varepsilon_{0}\) and \(\sigma\) are constants. The force corresponding to this potential is \(F(r)=-V^{\prime}(r) .\) Find \(F(r)\).

Electric Potential Suppose that a ring-shaped conductor of radius \(a\) carries a total charge \(Q\). Then the electrical potential at the point \(P\), a distance \(x\) from the center and along the line perpendicular to the plane of the ring through its center, is given by $$ V(x)=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{\sqrt{x^{2}+a^{2}}} $$ where \(\varepsilon_{0}\) is a constant called the permittivity of free space. The magnitude of the electric field induced by the charge at the point \(P\) is \(E=-d V / d x\), and the direction of the field is along the \(x\) -axis. Find \(E\).

Simple Harmonic Motion The equation of motion of a body executing simple harmonic motion is given by $$ x(t)=A \sin (\omega t+\phi) $$ where \(x\) (in feet) is the displacement of the body, \(A\) is the amplitude, \(\omega=\sqrt{k / m}, k\) is a constant, and \(m\) (in slugs) is the mass of the body. Find expressions for the velocity and acceleration of the body at time \(t\).

Two curves are said to be orthogonal if their tangent lines are perpendicular at each point of intersection of the curves. In Exercises \(89-92\), show that the curves with the given equations are orthogonal.$$ y-x=\frac{\pi}{2}, \quad x=\cos y $$

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.