Problem 24
Question
Find the derivative. Assume \(a, b, c, k\) are constants. $$y=z^{2}+\frac{1}{2 z}$$
Step-by-Step Solution
Verified Answer
The derivative of the function is \(\frac{dy}{dz} = 2z - \frac{1}{2z^2}\).
1Step 1: Differentiate the First Term
The first term of the function is \(z^2\). Use the power rule for differentiation, which states that \( \frac{d}{dz}(z^n) = nz^{n-1} \). Here, \( n = 2 \), so the derivative of \( z^2 \) is \( 2z \).
2Step 2: Differentiate the Second Term
The second term of the function is \( \frac{1}{2z} \). Rewrite it as \( \frac{1}{2} z^{-1} \). Use the power rule for differentiation. Here, \( n = -1 \), so the derivative is \( \frac{1}{2} (-1)z^{-2} = -\frac{1}{2z^2} \).
3Step 3: Combine the Derivatives
Add the derivatives from Step 1 and Step 2 to find the derivative of the entire function. The result is \( \frac{dy}{dz} = 2z - \frac{1}{2z^2} \). This represents the derivative of \( y \) with respect to \( z \).
Key Concepts
Power Rule for DifferentiationConstants in DifferentiationDerivative Calculation Process
Power Rule for Differentiation
Differentiating functions might seem intimidating, but the power rule makes it much easier. The power rule is a fundamental tool in calculus, allowing us to find the derivative of functions with ease. In simple terms, it describes how to differentiate expressions like \(z^n\), where \(n\) is any constant exponent.
The rule states that to find the derivative of \(z^n\), you multiply the entire expression by the exponent \(n\) and then reduce the exponent by one. Mathematically, it's represented as:
For example, if you have a term \(z^2\), as in our original exercise, applying the power rule gives you \(2z^{2-1}\) or simply \(2z\).
The simplicity and power of the power rule make it an essential element in calculus, and it is applicable to polynomials and any other functions where each term is a variable raised to a constant power.
The rule states that to find the derivative of \(z^n\), you multiply the entire expression by the exponent \(n\) and then reduce the exponent by one. Mathematically, it's represented as:
- \(\frac{d}{dz}(z^n) = nz^{n-1}\)
For example, if you have a term \(z^2\), as in our original exercise, applying the power rule gives you \(2z^{2-1}\) or simply \(2z\).
The simplicity and power of the power rule make it an essential element in calculus, and it is applicable to polynomials and any other functions where each term is a variable raised to a constant power.
Constants in Differentiation
When working with differentiation, constants play a straightforward role. They are terms or coefficients in a function that do not involve variables. Understanding how to treat constants is crucial when differentiating composite expressions.
Firstly, constants directly multiplying the variable just stay the same. For instance, if you have a function like \(3z^3\), the constant is \(3\) and it remains unchanged when applying the power rule.
Similarly, the derivative of a constant term without any variable involvement whatsoever is zero. Essentially, constants only act as scalable factors for terms with variables:
In our exercise, the term \(\frac{1}{2z}\) was rewritten as \(\frac{1}{2}z^{-1}\), where \(\frac{1}{2}\) is a constant. It scales the derivative according to the power rule while remaining unchanged through the differentiation process. This flexibility in handling constants simplifies the calculation of derivatives in most functions.
Firstly, constants directly multiplying the variable just stay the same. For instance, if you have a function like \(3z^3\), the constant is \(3\) and it remains unchanged when applying the power rule.
Similarly, the derivative of a constant term without any variable involvement whatsoever is zero. Essentially, constants only act as scalable factors for terms with variables:
- \(\frac{d}{dz}(c \cdot z^n) = c \cdot nz^{n-1}\)
In our exercise, the term \(\frac{1}{2z}\) was rewritten as \(\frac{1}{2}z^{-1}\), where \(\frac{1}{2}\) is a constant. It scales the derivative according to the power rule while remaining unchanged through the differentiation process. This flexibility in handling constants simplifies the calculation of derivatives in most functions.
Derivative Calculation Process
Calculating derivatives is about breaking down a function into smaller, manageable parts and applying differentiation rules accurately. From there, you combine those differentiated parts back together to find the derivative of the whole function.
Let's take another look at how we approached this in our example:
1. **Differentiate Each Term:** Start by applying the power rule we discussed. For any term like \(z^n\), differentiate it using \(nz^{n-1}\). In our example, for \(z^2\), it became \(2z\).
2. **Deal with Constants:** Recognize and keep constants as they multiply the derived variables unless they are standalone terms, in which case their derivative is zero.
3. **Combine All Results:** Once you've differentiated each term, sum them up to get the derivative of the original function.
In the end, the derivative of the function \(y = z^2 + \frac{1}{2z}\) is shown as \(\frac{dy}{dz} = 2z - \frac{1}{2z^2}\). Calculating this involved applying the differentiation rules systematically and combining them to achieve the result.
Practicing this step-by-step approach will strengthen your understanding and make the process more intuitive over time.
Let's take another look at how we approached this in our example:
1. **Differentiate Each Term:** Start by applying the power rule we discussed. For any term like \(z^n\), differentiate it using \(nz^{n-1}\). In our example, for \(z^2\), it became \(2z\).
2. **Deal with Constants:** Recognize and keep constants as they multiply the derived variables unless they are standalone terms, in which case their derivative is zero.
3. **Combine All Results:** Once you've differentiated each term, sum them up to get the derivative of the original function.
In the end, the derivative of the function \(y = z^2 + \frac{1}{2z}\) is shown as \(\frac{dy}{dz} = 2z - \frac{1}{2z^2}\). Calculating this involved applying the differentiation rules systematically and combining them to achieve the result.
Practicing this step-by-step approach will strengthen your understanding and make the process more intuitive over time.
Other exercises in this chapter
Problem 24
Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ w=\frac{3 z}{1+2 z} $$
View solution Problem 24
Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(D=10-\ln p\)
View solution Problem 25
A company's monthly sales, \(S(t)\), are seasonal and given as a function of time, \(t\), in months, by $$ S(t)=2000+600 \sin \left(\frac{\pi}{6} t\right) . $$
View solution Problem 25
Find the derivative. Assume that \(a, b, c\), and \(k\) are constants. $$ z=\frac{1-t}{1+t} $$
View solution