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Differentiate. $$ y=x e^{-2 x}+e^{-x}+x^{3} $$

Short Answer

Expert verified
The derivative is \(e^{-2x} - 2xe^{-2x} - e^{-x} + 3x^2\).

Step by step solution

01

Differentiate First Term

The first term is \(xe^{-2x}\). Use the product rule to differentiate it, as it is a product of \(x\) and \(e^{-2x}\). The product rule says that if \(u(x)\) and \(v(x)\) are functions, then \((uv)' = u'v + uv'\). Here, let \(u = x\) and \(v = e^{-2x}\) so that \(u' = 1\) and \(v' = -2e^{-2x}\). Therefore, the derivative of the first term is \(1 imes e^{-2x} + x imes (-2e^{-2x}) = e^{-2x} - 2xe^{-2x}\).
02

Differentiate Second Term

The second term is \(e^{-x}\). Differentiate this exponential function by using the chain rule. The derivative of \(e^{u}\) is \(e^{u}\times u'\). Here \(u = -x\), so the derivative is \(-e^{-x}\).
03

Differentiate Third Term

The third term is \(x^3\). Differentiate using the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\). Applying the power rule, the derivative of \(x^3\) is \(3x^2\).
04

Combine Derivatives

Combine the derivatives of each term to find the derivative of the entire function. This gives us \((e^{-2x} - 2xe^{-2x}) + (-e^{-x}) + 3x^2\). Simplify the expression to obtain the final result.

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Key Concepts

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

Product Rule
When you're faced with differentiating a function that is the product of two smaller functions, the product rule is your best friend. The product rule states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product is not simply the product of their derivatives. Instead, it's given by:
  • \((uv)' = u'v + uv'\)
This means you take the derivative of the first function \( (u') \) and multiply it by the second function \( (v) \), then add the first function \( (u) \) multiplied by the derivative of the second function \( (v') \). This rule is particularly useful when dealing with terms like \( xe^{-2x} \). Here, \( u = x \) and \( v = e^{-2x} \), meaning \( u' = 1 \) and \( v' = -2e^{-2x} \). Applying the product rule, we find:\[ 1 \cdot e^{-2x} + x \cdot (-2e^{-2x}) = e^{-2x} - 2xe^{-2x}. \] By using this rule, you can easily tackle any products of functions in differentiation tasks.
Chain Rule
The chain rule is a powerful tool in calculus used to differentiate composite functions. Think of it like peeling an onion; you have layers of functions and need to deal with each layer at a time. The chain rule tells us that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function. In mathematical terms:
  • If \( y = f(g(x)) \) then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)
For example, if you need to differentiate a function like \( e^{-x} \), you use the chain rule. Here, the outer function is \( e^u \) and the inner function is \( u = -x \). Therefore, the derivative becomes:\[ f'(g(x)) = e^{-x} \quad \text{and} \quad g'(x) = -1, \]leading to the derivative of the term being:\[ -e^{-x}. \]Whenever you see a function nested within another, the chain rule is the key to finding the derivative correctly.
Power Rule
The power rule is perhaps the simplest and most frequently used rule in differentiation. It provides a quick and straightforward method to differentiate functions of the form \( x^n \). According to the power rule, the derivative of \( x^n \) is:
  • \( \frac{d}{dx}x^n = nx^{n-1}\)
This means you multiply the current exponent by the coefficient and subtract one from the exponent. For instance, differentiating \( x^3 \) gives us:\[ 3x^{3-1} = 3x^2. \]This rule applies to any real number values of \( n \). It's a fundamental rule that helps simplify the process of finding derivatives of polynomial functions quickly.

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Most popular questions from this chapter

Differentiate. $$ f(t)=\frac{\ln t^{2}}{t^{2}} $$

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