Problem 31 - Applied Calculus

Question

Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ f(x)=a x e^{-b x} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = a e^{-b x} (1 - b x) \).

Step 1: Identify the Functions

The given function is a product of two simpler functions: \( a x \) and \( e^{-b x} \). We will use the product rule for differentiation.

Step 2: Apply the Product Rule Formula

The product rule for derivatives is \( (uv)' = u'v + uv' \) where \( u = a x \) and \( v = e^{-b x} \). Compute these derivatives separately.

Step 3: Differentiate \( u = a x \)

The derivative of \( u = a x \) with respect to \( x \) is \( u' = a \), because the derivative of \( x \) is 1 and \( a \) is a constant.

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

The derivative of \( v = e^{-b x} \) with respect to \( x \) is \( v' = -b e^{-b x} \). This follows from the chain rule where the derivative of \( e^x \) is itself and we multiply by the derivative of \(-b x\), which is \(-b\).

Step 5: Substitute into the Product Rule Formula

Substitute \( u' = a \), \( u = a x \), \( v' = -b e^{-b x} \), and \( v = e^{-b x} \) into the product rule: \[(a x e^{-b x})' = a e^{-b x} + a x (-b e^{-b x})\].

Step 6: Simplify the Expression

Simplify the expression: \( a e^{-b x} - a b x e^{-b x} \). Factor out \( a e^{-b x} \):\[(a x e^{-b x})' = a e^{-b x} (1 - b x)\].

Key Concepts

Product Rule

The product rule is a fundamental principle in calculus. It helps us differentiate functions that are multiplied together. For a product of two functions, say \( u(x) \) and \( v(x) \), the product rule formula states:

• \((uv)' = u'v + uv'\)

This means that to find the derivative of the product, we must differentiate each function separately. Once done, we multiply the derivative of the first function by the second function as it is. Next, we repeat the process but reverse the roles. Finally, we sum these products together.
In our case, the function \( f(x) = a x e^{-b x} \) applies this rule. We first differentiate \( a x \) to get \( a \) and multiply it by \( e^{-b x} \). Then we differentiate \( e^{-b x} \) and multiply it by \( a x \). Adding these results gives us the full derivative.

Chain Rule

The chain rule is a crucial tool when dealing with composite functions, helping us find derivatives when functions are nested together. For a function \( y = g(f(x)) \), the chain rule is expressed as:

• \( \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \)

This formula essentially derives the outer function while keeping the inner function unchanged, then multiplies by the derivative of the inner function.
In the original problem, applying the chain rule to \( e^{-b x} \) involves treating the exponent \(-b x\) as the inner function. Differentiating \( e^{u} \) (where \( u \) is a function of \( x \)) gives \( e^{u} \cdot \frac{du}{dx} \). Therefore, the derivative of \( e^{-b x} \) becomes \( -b e^{-b x} \), since the derivative of \(-b x\) is \(-b\). This application is pivotal to completing the derivative calculation for our product rule exercise.

Exponential Functions

Exponential functions are defined in mathematics by the formula \( f(x) = a^x \), where \( a \) is a constant. A special case is the natural exponential function \( e^x \), where \( e \) is the base of natural logarithms, approximately 2.718. This specific function holds a unique property: its derivative is the same as the function itself.
Expanding to exponential functions such as \( e^{-b x} \), we employ the chain rule for differentiation. The function retains its original form but requires us to multiply by the derivative of the exponent. This property allows for exponential functions to change rapidly and remain prominent in numerous fields like finance and sciences.
In the function \( f(x) = a x e^{-b x} \), the presence of an exponential function makes the derivative calculation elegant, preserving the exponential term \( e^{-b x} \) even after differentiation. This allows for mathematical expressions to be easily managed and simplified, such as in our solution where we factor out \( a e^{-b x} \).