Question

Compute the derivative of the following functions. $$f(x)=(1-2 x) e^{-x}$$

Short answer

Answer: The derivative of the function is \(f'(x) = (1-2x)e^{-x} - 2e^{-x}\).

Step-by-step solution

Find the derivative of u(x)

First, let's find the derivative of u(x): $$u'(x) = \frac{d}{dx}(1-2x)$$ Using the basic differentiation rules, we get: $$u'(x) = -2$$

Find the derivative of v(x)

Next, let's find the derivative of v(x): $$v'(x) = \frac{d}{dx}(e^{-x})$$ Using the chain rule, we get: $$v'(x) = -e^{-x}$$

Apply the product rule

Now that we have u'(x) and v'(x), let's apply the product rule to find the derivative of f(x): $$f'(x) = u'(x)v(x)+u(x)v'(x)$$ Substituting the values we found in Steps 1 and 2, we get: $$f'(x) = (-2)e^{-x} + (1-2x)(-e^{-x})$$

Simplify the derivative

Finally, let's simplify the expression for f'(x): $$f'(x) = -2e^{-x} + e^{-x} - 2xe^{-x}$$ Factoring out the common term, \(e^{-x}\), we get: $$f'(x) = e^{-x}(-2+1-2x)$$ So the derivative of the function f(x) is: $$f'(x) = (1-2x)e^{-x} - 2e^{-x}$$

Key concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It's a fundamental concept needed to understand rates of change, and knowing how to compute derivatives is crucial. For example, if you have a function like \(f(x) = (1-2x)e^{-x}\), finding its derivative tells us about the rate at which \(f(x)\) changes with \(x\). By employing basic differentiation rules, you can determine this rate effectively. Differentiation helps us understand the underlying behavior of the function, such as whether it is increasing, decreasing, or remaining constant within its domain.

Product Rule

The product rule is a technique used when differentiating functions that are the multiplication of two dependent functions. For instance, with the function \(f(x) = u(x)v(x)\), where \(u(x) = (1-2x)\) and \(v(x) = e^{-x}\), the product rule states that:

• The derivative \(f'(x)\) is given by \(f'(x) = u'(x)v(x) + u(x)v'(x)\).
• This basically involves taking the derivative of the first function and multiplying it by the second function, then adding the product of the first function and the derivative of the second function.

This rule helps break down complex functions into manageable parts, enabling easier differentiation.

Chain Rule

The chain rule is a vital tool in calculus for finding the derivative of composite functions. A composite function is one where a function is inside another function, like \(v(x) = e^{-x}\) in our example. To differentiate \(v(x)\), we need the chain rule because \(e^{-x}\) essentially is a function \(e^u\) where \(u = -x\). Here's how it works:

• Differentiate the outer function, \(e^u\), with respect to \(u\), which gives us \(e^u\).
• Then, differentiate the inner function \(-x\) with respect to \(x\), yielding \(-1\).
• Finally, multiply these derivatives together to get \(-e^{-x}\).

Ultimately, the chain rule simplifies the process of tackling more complicated derivatives involving nested functions.

Differentiation

Differentiation is the process of finding a derivative and is pivotal in calculus. Whether you're dealing with simple polynomials or more complicated expressions involving products and compositions of functions, the concepts of differentiation help break down the problem. With our initial function \(f(x) = (1-2x)e^{-x}\), differentiation involves using rules like the product rule and the chain rule. By applying these rules correctly, we find that:

• The derivative of \(u(x) = 1-2x\) is \(-2\).
• The derivative of \(v(x) = e^{-x}\) is \(-e^{-x}\).
• Combining these results with the product rule gives us the simplified expression for the derivative.

Differentiation is essential for exploring the unique features and behavior of functions, enabling us to compute areas, optimize problems, and understand motion.