Problem 11 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=x^{2} e^{-x} $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = x^2 e^{-x} \) is \( f'(x) = e^{-x}(2x - x^2) \).

Step 1: Identify the type of function

The function given is a product of two functions. It is of the form \( f(x) = u(x) \, v(x) = x^2 \, e^{-x} \), where \( u(x) = x^2 \) and \( v(x) = e^{-x} \). This suggests that we should use the product rule for differentiation.

Step 2: Apply the product rule

The product rule states that the derivative of a product \( u(x) \, v(x) \) is given by \( u'(x) \, v(x) + u(x) \, v'(x) \). We need to calculate \( u'(x) \) and \( v'(x) \) first before applying the rule.

Step 3: Differentiate \( u(x) = x^2 \)

The derivative of \( u(x) = x^2 \) with respect to \( x \) is \( u'(x) = 2x \), using the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \).

Step 4: Differentiate \( v(x) = e^{-x} \)

The function \( v(x) = e^{-x} \) is an exponential function. Its derivative is \( v'(x) = -e^{-x} \), using the rule \( \frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x) \) where \( g(x) = -x \) and \( g'(x) = -1 \).

Step 5: Substitute into the product rule

Using the product rule, substitute \( u(x) = x^2 \), \( u'(x) = 2x \), \( v(x) = e^{-x} \), and \( v'(x) = -e^{-x} \) into the formula: \( f'(x) = u'(x) \, v(x) + u(x) \, v'(x) = 2x \, e^{-x} + x^2 \, (-e^{-x}) \).

Step 6: Simplify the expression

Combine like terms: \( f'(x) = 2x \, e^{-x} - x^2 \, e^{-x} \). Factoring out the common term \( e^{-x} \), we have \( f'(x) = e^{-x} (2x - x^2) \).

Key Concepts

Exponential Function Differentiation

In calculus, differentiating exponential functions involves recognizing the distinctive characteristic of the exponential function itself.
The exponential function, particularly when it has the base of Euler's number, allows for a straightforward rule.
If you have a function of the form \(e^{g(x)}\), its derivative is found by multiplying the original function by the derivative of the exponent:

• Derivative formula: \(\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)\)
• In our example, \(v(x) = e^{-x}\), and \(g(x) = -x\), where the derivative \(g'(x) = -1\).

This gives the derivative \(v'(x) = -e^{-x}\). Recognizing and applying this rule is crucial in handling exponential functions in differentiation.

Power Rule Differentiation

The power rule is one of the most fundamental differentiation rules.
It makes the process of finding derivatives for functions involving powers of \(x\) much easier to handle.
The rule states:

• If you have a function of the form \(x^n\), the derivative is \(nx^{n-1}\).
• In the example we've been given, \(u(x) = x^2\), so the derivative \(u'(x) = 2x\).

The power rule is essential because it simplifies working with polynomial expressions, making differentiation quick and efficient. Always apply this rule to each term individually when faced with polynomials or products.

Derivative Simplification

After differentiating individual components, the next step often involves combining and simplifying the expression.
This is critical for providing a clear and concise representation of the derivative.
To simplify:

• Combine like terms in the expression.
• Factor out common terms when possible.

In our example, the unsimplified derivative was \(f'(x) = 2x \, e^{-x} - x^2 \, e^{-x}\).
By recognizing \(e^{-x}\) as a common factor, it allows us to express the derivative more neatly as \(f'(x) = e^{-x} (2x - x^2)\).
Simplification enhances clarity and aids in further applications of the derivative.

Factoring in Calculus

Factoring is a key tool in calculus for simplifying expressions and solving equations.
After differentiating, factoring can reveal more about the function's behavior, such as finding zeroes or simplifying integrals and derivatives.
When factoring:

• Identify common factors across terms in your expression.
• Factor out the greatest common factor to simplify as much as possible.
• A simplified function can make further calculations or interpretations easier.

In our function \(f'(x) = 2x \, e^{-x} - x^2 \, e^{-x}\), factoring out \(e^{-x}\) not only simplifies the expression to \(e^{-x} (2x - x^2)\) but also provides a factorized form that may be more insightful for certain calculus applications, like analyzing the function's graph or finding local extrema.