/*! 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 90 Differentiate. $$y=\left(e^{x^... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate. $$y=\left(e^{x^{2}}-2\right)^{4}$$

Short Answer

Expert verified
\(\frac{dy}{dx} = 8x(e^{x^2} - 2)^3 e^{x^2}\)

Step by step solution

01

Recognize the Outer Function

The given function is a composite function. Identify the outer function as \(u = (e^{x^2} - 2)^4\).
02

Differentiate the Outer Function

Using the chain rule, differentiate the outer function \(u = v^4\) where \(v = e^{x^2} - 2\). The derivative is \[\frac{d}{dv}(v^4) = 4v^3\].
03

Identify and Differentiate the Inner Function

The inner function is \(v = e^{x^2} - 2\). The derivative of the inner function using the chain rule is \[\frac{d}{dx}(e^{x^2} - 2) = \frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x = 2xe^{x^2}\].
04

Apply the Chain Rule

Combine the results using the chain rule: \[(u)' = \frac{d}{dx}((e^{x^2} - 2)^4) = 4(e^{x^2} - 2)^3 \cdot \frac{d}{dx}(e^{x^2} - 2) = 4(e^{x^2} - 2)^3 \cdot 2xe^{x^2} = 8x(e^{x^2} - 2)^3 e^{x^2}\]
05

Combine the Results

Combining everything together, we get the final derivative:\(\frac{dy}{dx} = 8x(e^{x^2} - 2)^3 e^{x^2}\).

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 91Ó°ÊÓ!

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 for finding the derivative of a composite function. A composite function is essentially a function within another function, similar to nesting one matryoshka doll inside another. When we can't directly differentiate a function, the chain rule allows us to do so by breaking it down into simpler, distinguishable parts.
The chain rule states that if you have a composite function, say \(y = f(g(x))\), then its derivative can be found by differentiating the outer function \(f\) with respect to the inner function \(g\) and then multiplying it by the derivative of the inner function \(g\). Mathematically, the chain rule is expressed as:
\[\frac{dy}{dx} = \frac{d}{dg(x)}f(g(x)) \cdot \frac{d}{dx}g(x)\]
This rule becomes very handy when dealing with complex expressions that cannot be easily simplified.
composite functions
A composite function involves two or more functions combined, where one function's output becomes the input of another. For example, if you have functions \(f(x)\) and \(g(x)\), a composite function can be written as \(f(g(x))\). This can be visualized like a production line where each station (function) processes something before passing it on to the next.
In our exercise, the function \(y=(e^{x^2} - 2)^4\) is a composite function. The outer function is \(u(x) = (v)^4\), and the inner function is \(v(x) = e^{x^2} - 2\). To differentiate such functions, you first focus on the outer function and then work your way inward. This approach allows the chain rule to simplify and solve the problem section by section.
differentiation
Differentiation is the process of finding the derivative of a function. The derivative represents how the function's output changes as its input changes—in other words, it's a measure of the function's rate of change. For any given function \(f(x)\), the derivative, denoted as \(f'(x)\) or \(\frac{df}{dx}\), gives us important insights like the slope of the function at any point.
In our step-by-step solution, we use differentiation to find the slope of the composite function \(y=(e^{x^2} - 2)^4\). By differentiating both the outer function \( (v)^4 \) and the inner function \( e^{x^2} - 2 \), and then combining these using the chain rule, we achieve the final derivative:
\[\frac{dy}{dx} = 8x(e^{x^2} - 2)^3 e^{x^2}\]
This demonstrates the power of differentiation in breaking down and understanding the behavior of 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

Find the equation of the line tangent to the graph of \(y=\left(x^{2}-x\right) \ln (6 x)\) at \(x=2\)

Differentiate. $$y=5^{2 x^{3}-1} \cdot \log (6 x+5)$$

Students in a zoology class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After \(t\) months, the average score \(S(t),\) as a percentage, was found to be given by \(S(t)=78-15 \ln (t+1), \quad t \geq 0\) a) What was the average score when they initially took the test, \(t=0 ?\) b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years \((24 \text { months }) ?\) e) Find \(S^{\prime}(t)\) f) Find the maximum and minimum values, if they exist. g) Find \(\lim _{t \rightarrow \infty} S(t)\) and discuss its meaning.

Differentiate. $$f(x)=\frac{1}{5} x^{5}\left(\ln x-\frac{1}{5}\right)$$

Students in a botany class took a final exam. They took equivalent forms of the exam at monthly intervals thereafter. After \(t\) months, the average score \(S(t),\) as a percentage, was found to be \(S(t)=68-20 \ln (t+1), \quad t \geq 0\) a) What was the average score when the students initially took the test? b) What was the average score after 4 months? c) What was the average score after 24 months? d) What percentage of their original answers did the students retain after 2 years ( 24 months)? e) Find \(S^{\prime}(t)\) f) Find the maximum value, if one exists. g) Find \(\lim _{t \rightarrow \infty} S(t),\) and discuss its meaning.

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.