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Chapter 3: Problem 33

Find the derivatives of the function. $$y=x^{9 / 4}+e^{-2 x}$$

Short Answer

Expert verified
\(\frac{dy}{dx} = \frac{9}{4}x^{5/4} - 2e^{-2x}\)

Step by step solution

01

Differentiate the First Term

The first term of the function is \(x^{9/4}\). Use the power rule for differentiation, which states that if \(y = x^n\), then \(\frac{dy}{dx} = nx^{n-1}\). For \(x^{9/4}\), differentiate it to get \(\frac{9}{4}x^{9/4 - 1} = \frac{9}{4}x^{5/4}\).
02

Differentiate the Second Term

The second term of the function is \(e^{-2x}\). Use the chain rule for differentiation. The derivative of \(e^{f(x)}\) is \(f'(x)e^{f(x)}\). Here, \(f(x)=-2x\), and \(f'(x)=-2\). The derivative is \(-2e^{-2x}\).
03

Combine the Derivatives

Now combine the derivatives of both terms. The derivative of the entire function \(y = x^{9/4} + e^{-2x}\) is the sum of the derivatives found in the previous steps. So, the derivative \(\frac{dy}{dx} = \frac{9}{4}x^{5/4} - 2e^{-2x}\).

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Key Concepts

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

Power Rule
The power rule is a fundamental tool in differentiation used to find the derivative of functions of the form \(x^n\). This rule simplifies the process by guiding us to "bring down" the exponent as a coefficient and then reduce the original exponent by one.
  • If you have a function \(y = x^n\), its derivative is \(\frac{dy}{dx} = nx^{n-1}\).
  • For example, differentiating \(x^{9/4}\) using the power rule results in \(\frac{9}{4}x^{9/4 - 1} = \frac{9}{4}x^{5/4}\).
This process highlights the elegance of the power rule, making it a preferred choice for differentiating polynomial functions, simplifying otherwise complex expressions.
Chain Rule
The chain rule is an essential method in calculus for differentiating composite functions. It allows you to break down complicated expressions and differentiate them step by step.
  • Consider a function \(y = g(f(x))\), where you must first differentiate the outer function, \(g\), with respect to the inner function, \(f\).
  • The chain rule formula is \( (g(f(x)))' = g'(f(x))\cdot f'(x) \).
  • In our example, \(e^{-2x}\), the challenges are tackled by first identifying \(f(x) = -2x\) and \(g(u) = e^u\), resulting in a derivative of \(-2e^{-2x}\).
The chain rule facilitates differentiation when you have nested functions, providing a powerful tool to handle complex mathematical problems.
Differentiation
Differentiation is the cornerstone of calculus, aimed at finding the rate at which a function changes at any given point. This process is fundamental in understanding and analyzing mathematical models of real-world situations.
  • At its core, differentiation involves applying rules, like the power rule and chain rule, to compute derivatives.
  • The derivative of a function, represented as \(\frac{dy}{dx}\), provides insight into the function's instantaneous rate of change.
  • For the function \(y = x^{9/4} + e^{-2x}\), we differentiate each term separately and combine the results, obtaining \(\frac{9}{4}x^{5/4} - 2e^{-2x}\).
This foundational skill has vast applications in physics, engineering, economics, and beyond, making it crucial for students to grasp and apply various differentiation techniques.
Exponential Function
Exponential functions are unique in calculus because they involve constants raised to a variable power, often appearing in the form \(e^{f(x)}\). These functions have distinct rules for differentiation that require careful application of techniques like the chain rule.
  • The exponential function \(e^x\) has a unique property where its derivative is itself, \(e^x\).
  • However, for \(e^{f(x)}\), we need to apply the chain rule: differentiate \(f(x)\) first and then multiply by the exponential term.
  • In the problem example, \(e^{-2x}\) demands this method, as we must also differentiate the inner function \(-2x\), resulting in \(-2e^{-2x}\).
Understanding the behavior and differentiation of exponential functions opens up a world of modeling growth, decay, and many natural processes, demonstrating their significance in various fields.

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Most popular questions from this chapter

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