Problem 21 - Single Variable Calculus

Question

Find the derivative of each function. . \(f(x)=\frac{e^{x}(x+1)}{x-2}\)

Step-by-Step Solution

Answer

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

Step 1: Identify u(x) and v(x)

We can identify \(u(x) = e^x(x+1)\) and \(v(x) = x-2\).

Step 2: Find derivative of u(x) and v(x)

Now we need to find the derivatives of u(x) and v(x), i.e., \(u'(x)\) and \(v'(x)\). By applying the product rule for derivatives, which states that \((uv)' = u'v + uv'\), we get: \(u'(x) = \frac{d}{dx}( e^x (x+1)) = e^x (x+1) + e^x\) \(v'(x) = \frac{d}{dx}(x-2) = 1\)

Step 3: Apply Quotient Rule for Derivatives

Now we apply the quotient rule: \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). Substitute the values of \(u(x), u'(x), v(x),\) and \(v'(x)\) in the above equation to get: \(f'(x) = \frac{(e^x (x+1) + e^x)(x-2) - e^x(x+1)}{(x-2)^2}\)

Step 4: Simplify the Expression

We can simplify the equation further to obtain the final derivative: \(f'(x) = \frac{e^x (x+1)(x-2) + e^x(x-2) - e^x(x+1)}{(x-2)^2}\) Combine like terms: \(f'(x) = \frac{e^x (x-1)}{(x-2)^2}\) The derivative of the given function is \(f'(x) = \frac{e^x (x-1)}{(x-2)^2}\).

Key Concepts

Quotient Rule

In calculus, the Quotient Rule provides a way to differentiate functions that are expressed as a quotient of two other functions. If you have a function that can be written as \(f(x) = \frac{u(x)}{v(x)}\), where both \(u(x)\) and \(v(x)\) are differentiable, the Quotient Rule helps you find the derivative, \(f'(x)\). The rule is stated as follows: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \] This process involves the derivative of the numerator \(u(x)\), known as \(u'(x)\), and the derivative of the denominator \(v(x)\), or \(v'(x)\). You multiply the derivative of the numerator by the original denominator and then subtract the product of the original numerator and the derivative of the denominator. Finally, you divide the result by the square of the original denominator \((v(x))^2\). Performing these steps carefully allows you to find the derivative of the quotient accurately.

Product Rule

The Product Rule is essential when differentiating two functions that are multiplied together. Given two differentiable functions, \(u(x)\) and \(v(x)\), the derivative of their product is expressed with the Product Rule: \[ (uv)' = u'v + uv' \] Here, you calculate the derivative of the first function \(u(x)\) and then multiply it by the second function \(v(x)\). Similarly, take the original function \(u(x)\) and multiply it by the derivative of \(v(x)\). The sum of these two products gives the derivative of the product of these functions. This rule was key in differentiating the function \(u(x) = e^x(x+1)\) in the original exercise. The proper application of this rule eases the computation and reduces errors in complex expressions.

Derivative Simplification

After applying differentiation rules, the resulting expression can often be complex. Simplification involves combining like terms, simplifying fractions, and reducing expressions to their simplest form. This process is crucial as it makes the result more readable and manageable.

• Combine terms that have the same base and exponent.
• Factor common terms whenever possible to simplify fractions.
• Reduce the expression by canceling out terms where applicable.

In our problem, the simplification helped transform a complicated derivative expression into a more elegant form: \(f'(x) = \frac{e^x (x-1)}{(x-2)^2}\). Although the expression initially looked daunting, systematic simplification made the solution clearer.

Exponential Functions

Exponential functions are a significant topic in calculus due to their unique properties and frequent occurrence in applications. The general form of an exponential function is \(y = e^{x}\), where \(e\) is Euler's number, approximately 2.718. One remarkable feature of exponential functions is that their derivatives are proportional to the original function.

• The derivative of \(e^x\) is \(e^x\).
• For a function \(e^{u(x)}\), where \(u(x)\) is diffentiable, the chain rule gives \(\frac{d}{dx} e^{u(x)} = e^{u(x)} u'(x)\).

In the context of our exercise, \(e^x\) appears in the multiplied functional form \(e^x(x+1)\). Consequently, understanding exponential functions' behaviors and differentiating them is vital for handling such expressions efficiently.