Question

Derivatives Find and simplify the derivative of the following functions. $$f(z)=\left(\frac{z^{2}+1}{z}\right) e^{z}$$

Short answer

Question: Find the simplified derivative of the function \(f(z) = \frac{z^2 + 1}{z}e^z\). Answer: The simplified derivative of the function \(f(z)\) is \(f'(z) = e^z\left(\frac{z^3 + z^2 - 1}{z^3}\right)\).

Step-by-step solution

Identify the functions and their derivatives

We can rewrite the function \(f(z)\) as a product of two functions like this: $$f(z) = g(z) \cdot h(z),$$ where \(g(z) = \frac{z^2 + 1}{z}\) and \(h(z) = e^z\). We will first find the derivatives of \(g(z)\) and \(h(z)\) separately.

Find the derivative of g(z) using the quotient rule

The quotient rule states that for a function \(g(z) = \frac{u(z)}{v(z)}\), its derivative is given by $$g'(z) = \frac{u'(z)v(z) - u(z)v'(z)}{[v(z)]^2}.$$ Here, we have \(u(z) = z^2 + 1\) and \(v(z) = z\). Therefore, we have \(u'(z) = 2z\) and \(v'(z) = 1\). Applying the quotient rule: $$g'(z) = \frac{(2z)(z) - (z^2 + 1)(1)}{z^2} = \frac{2z^2 - z^2 - 1}{z^2}.$$

Find the derivative of h(z)

The function \(h(z) = e^z\) is an exponential function, and its derivative is given by: $$h'(z) = e^z.$$

Apply the product rule to find the derivative of f(z)

The product rule states that for a function \(f(z) = g(z) \cdot h(z)\), its derivative is given by: $$f'(z) = g'(z)h(z) + g(z)h'(z).$$ Now, we have \(g'(z) = \frac{2z^2 - z^2 - 1}{z^2}\), \(h(z) = e^z\), and \(h'(z) = e^z\). Substitute these expressions into the product rule: $$f'(z) = \left(\frac{2z^2 - z^2 - 1}{z^2}\right)e^z + \left(\frac{z^2 + 1}{z}\right)e^z.$$

Simplify the derivative of f(z)

To simplify the expression, factor out the common term \(e^z\): $$f'(z) = e^z\left(\frac{2z^2 - z^2 - 1}{z^2} + \frac{z^2 + 1}{z}\right).$$ Now, make both terms have a common denominator by multiplying the second term by \(z\): $$f'(z) = e^z\left(\frac{2z^2 - z^2 - 1}{z^2} + \frac{(z^2 + 1)z}{z^3}\right).$$ Add the fractions and simplify: $$f'(z) = e^z\left(\frac{2z^2 - z^2 - 1 + z^3 + z}{z^3}\right) = e^z\left(\frac{z^3 + z^2 - 1}{z^3}\right).$$ Thus, the simplified derivative of the function \(f(z)\) is: $$f'(z) = e^z\left(\frac{z^3 + z^2 - 1}{z^3}\right).$$

Key concepts

Quotient Rule

When dealing with the derivatives of functions that are fractions, we often use the quotient rule. It is crucial for solving complex derivatives where one function is divided by another.

The quotient rule formula is expressed as: \[\begin{equation}\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2},\end{equation}\]where \( u \) and \( v \) are differentiable functions of \( z \). Here, \( u' \) and \( v' \) stand for the derivatives of \( u \) and \( v \) respectively.

When applying the quotient rule, careful calculation is necessary to correctly find each part of the formula, especially when simplifying algebraic expressions. This helps in avoiding common mistakes, such as mixing up the order of subtraction in the numerator or squaring the denominator incorrectly.

For the exercise provided, the quotient rule is applied to the function \( g(z) \), helping to find its derivative. This step is crucial for simplifying the derivative of the overall function \( f(z) \) as it combines with the exponential part.

Product Rule

The product rule is another foundational tool in calculus, used when taking the derivative of a product of two or more functions. The rule may seem challenging at first, but it's simply about recognizing patterns and systematic application.

The product rule formula is:\[\begin{equation}(fg)' = f'g + fg',\end{equation}\]where \( f \) and \( g \) are functions whose derivatives \( f' \) and \( g' \) are known.

Applying this rule requires you to differentiate each function separately and then sum the products of each function and the derivative of the other. This step must be carried out with precision to ensure correct results.

In our given exercise, after using the quotient rule to find \( g'(z) \), the product rule is employed to combine the derivatives of \( g(z) \) and the exponential function \( h(z) = e^z \). It's a powerful demonstration of how these rules work in tandem to simplify complex derivatives.

Exponential Function Derivative

The derivative of an exponential function is a simpler concept compared to the product and quotient rules, yet it is equally important. Exponential functions have the basic form \( a^x \), where \( a \) is a constant, and have unique properties for their derivatives.

For the special case where the base is the natural logarithm base \( e \), the derivative of \( e^x \) is quite elegant: it is simply equal to \( e^x \). This property makes the exponential function very significant in both mathematics and applications involving growth and decay processes.

During our exercise with the function \( h(z) = e^z \), applying the derivative is straightforward since the function is already in the form of an exponential function with base \( e \). Its derivative is itself, which is a property exclusive to natural exponential functions. This simplicity is why exponential functions are often preferred in modeling real-world phenomena where rates of change are constantly proportional to the current value--as seen in interest rates, population growth, and radioactive decay.