Basic Rules of Differentiation: Videos & Practice Problems

In calculus, understanding how to find derivatives efficiently is crucial. Initially, derivatives can be calculated using limits, but there are simpler rules that can expedite the process. By learning a few fundamental rules, you can quickly determine the derivatives of various functions without resorting to limits.

One of the first rules to grasp is that the derivative of any constant function, such as \( g(x) = 7 \), is always zero. This is because a constant function has a slope of zero; there is no change in value regardless of the input. Therefore, we can express this mathematically as:

\[ \frac{d}{dx}(c) = 0 \] where \( c \) is any constant.

Next, consider the function \( f(x) = x \). The derivative of \( x \) is always one, reflecting that the slope of the tangent line to this linear function is constant and equal to one. This can be represented as:

\[ \frac{d}{dx}(x) = 1 \]

When combining functions, such as \( x + 7 \), we apply the sum and difference rule. This rule states that the derivative of the sum of two functions is the sum of their derivatives. Thus, for \( x + 7 \), we have:

\[ \frac{d}{dx}(x + 7) = \frac{d}{dx}(x) + \frac{d}{dx}(7) = 1 + 0 = 1 \]

Moving on to multiplication, the constant multiple rule is essential. For a function like \( 8x \), the derivative is not simply the product of the derivatives of 8 and \( x \). Instead, we factor out the constant and multiply it by the derivative of the function. Therefore:

\[ \frac{d}{dx}(8x) = 8 \cdot \frac{d}{dx}(x) = 8 \cdot 1 = 8 \]

When faced with more complex functions, such as \( 8x + 7 \), you can apply these rules collectively. By using the sum and constant multiple rules together, you can find the derivative efficiently. This approach will become second nature as you practice and apply these rules to various functions.

Understanding how to find derivatives is crucial in calculus, especially when dealing with polynomial functions. One of the fundamental tools for this is the power rule, which simplifies the process of differentiation for functions of the form \( f(x) = x^n \), where \( n \) is a real number.

The power rule states that the derivative of \( x^n \) can be calculated using the formula:

\( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \)

This means you multiply the function by the exponent \( n \) and then decrease the exponent by 1. For example, if you want to find the derivative of \( x^6 \), you would apply the power rule as follows:

\( \frac{d}{dx}(x^6) = 6 \cdot x^{6-1} = 6x^5 \)

When dealing with more complex functions that involve addition or subtraction, the sum rule comes into play. This rule allows you to differentiate each term separately. For instance, if you have a function \( f(x) = x^4 + x^3 \), you can find the derivative by applying the power rule to each term:

\( f'(x) = \frac{d}{dx}(x^4) + \frac{d}{dx}(x^3) = 4x^3 + 3x^2 \)

Additionally, when a function includes a constant multiplied by a variable, the constant multiple rule is useful. For example, for the function \( g(x) = 3x^2 \), you would first factor out the constant:

\( g'(x) = 3 \cdot \frac{d}{dx}(x^2) \)

Then, applying the power rule gives:

\( g'(x) = 3 \cdot 2x^{2-1} = 6x \)

By mastering the power rule, sum rule, and constant multiple rule, you will be well-equipped to tackle a wide range of polynomial functions and their derivatives. Continued practice will enhance your proficiency in differentiation, making it an essential skill as you progress in calculus.

The power rule is a fundamental tool in calculus for finding derivatives of power functions, including those with negative and rational exponents. The power rule states that if you have a function of the form \( f(x) = x^n \), the derivative \( f'(x) \) can be calculated using the formula:

\[ f'(x) = n \cdot x^{n-1} \]

For example, to find the derivative of \( f(x) = x^{-3} \), you apply the power rule by pulling the exponent out front and decreasing the exponent by one:

\[ f'(x) = -3 \cdot x^{-3-1} = -3 \cdot x^{-4} \]

This can also be expressed without a negative exponent as:

\[ f'(x) = -\frac{3}{x^4} \]

Next, consider the function \( f(x) = x^{\frac{3}{2}} \). Again, using the power rule:

\[ f'(x) = \frac{3}{2} \cdot x^{\frac{3}{2}-1} = \frac{3}{2} \cdot x^{\frac{1}{2}} \]

When dealing with functions that are not immediately in power form, such as \( f(x) = \frac{1}{x} \), it is helpful to rewrite them as power functions. Here, \( \frac{1}{x} \) can be rewritten as \( x^{-1} \). Applying the power rule gives:

\[ f'(x) = -1 \cdot x^{-1-1} = -1 \cdot x^{-2} = -\frac{1}{x^2} \]

For functions like \( f(x) = \sqrt[3]{x} \), which can be expressed as \( x^{\frac{1}{3}} \), the power rule can also be applied:

\[ f'(x) = \frac{1}{3} \cdot x^{\frac{1}{3}-1} = \frac{1}{3} \cdot x^{-\frac{2}{3}} \]

In summary, the power rule is versatile and can be applied to a wide range of functions, including those with negative and rational exponents. Mastering this rule is essential for effectively calculating derivatives in calculus.

To find the derivative of the function \( y = 2x^3 - x^2 + 4x + 7 \), we will apply the rules of differentiation step by step. The derivative is denoted as \( \frac{dy}{dx} \) or \( y' \), indicating that we are differentiating with respect to \( x \).

We can break down the function into individual terms and apply the sum and difference rules of derivatives. The function consists of four terms: \( 2x^3 \), \( -x^2 \), \( 4x \), and \( 7 \). We will differentiate each term separately:

1. For the first term \( 2x^3 \), we apply the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). Thus, the derivative is:

\[\frac{d}{dx}(2x^3) = 2 \cdot 3x^{3-1} = 6x^2\]

2. For the second term \( -x^2 \), we again use the power rule:

\[\frac{d}{dx}(-x^2) = -2x^{2-1} = -2x\]

3. The third term is \( 4x \). The derivative of \( x \) is \( 1 \), so:

\[\frac{d}{dx}(4x) = 4 \cdot 1 = 4\]

4. Finally, for the constant term \( 7 \), the derivative of any constant is \( 0 \):

\[\frac{d}{dx}(7) = 0\]

Now, we can combine all these derivatives together:

\[\frac{dy}{dx} = 6x^2 - 2x + 4 + 0\]

After simplifying, we find that the final derivative of the function is:

\[\frac{dy}{dx} = 6x^2 - 2x + 4\]

As you practice more with derivatives, you will become more proficient and may skip detailed steps, directly moving from the original function to its derivative. Continuous practice will enhance your understanding and speed in differentiation.

In this problem, we explore the function \( f(x) = 2x^3 - 3x + 5 \) and address several key concepts in calculus, including derivatives, tangent lines, and horizontal tangents.

To begin, we find the derivative of the function, denoted as \( f'(x) \). Utilizing the power rule, we differentiate each term of the polynomial. The derivative of \( 2x^3 \) is calculated by bringing down the exponent (3) and multiplying it by the coefficient (2), resulting in \( 6x^2 \). The next term, \( -3x \), simplifies to \( -3 \) since the derivative of \( x \) is 1. The constant term \( 5 \) has a derivative of 0. Therefore, the derivative of the function is:

\( f'(x) = 6x^2 - 3 \)

Next, we determine the equation of the tangent line at the point where \( x = 1 \). To do this, we need two key pieces of information: the slope of the tangent line (denoted as \( m \)) and a point on the line (denoted as \( (x_1, y_1) \)). We already know that \( x_1 = 1 \). To find \( y_1 \), we evaluate the original function at this point:

\( f(1) = 2(1)^3 - 3(1) + 5 = 2 - 3 + 5 = 4 \)

Thus, \( y_1 = 4 \). Now, we find the slope \( m \) by substituting \( x = 1 \) into the derivative:

\( f'(1) = 6(1)^2 - 3 = 6 - 3 = 3 \)

With \( m = 3 \) and the point \( (1, 4) \), we can use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), to write the equation of the tangent line:

\( y - 4 = 3(x - 1) \)

Distributing and simplifying gives:

\( y = 3x + 1 \)

Finally, we explore the conditions under which the tangent line is horizontal. A horizontal tangent line occurs when the slope is zero. Therefore, we set the derivative equal to zero:

\( 6x^2 - 3 = 0 \)

Solving for \( x \), we first add 3 to both sides:

\( 6x^2 = 3 \)

Next, we divide by 6:

\( x^2 = \frac{1}{2} \)

Taking the square root of both sides yields:

\( x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \)

In summary, we have derived the function's derivative, found the equation of the tangent line at \( x = 1 \), and identified the values of \( x \) where the tangent line is horizontal. This comprehensive approach reinforces the fundamental concepts of differentiation and the geometric interpretation of derivatives.

In this example, we explore the application of derivatives in a real-world context, specifically focusing on cell culture growth. The function representing the number of cells in a culture over time is given by p(t) = 1500 + 2t², where p is the number of cells and t is the time in hours.

To find the number of cells after 24 hours, we evaluate p(24):

p(24) = 1500 + 2(24)²

Calculating this gives:

p(24) = 1500 + 2(576) = 1500 + 1152 = 2652

This result indicates that after 24 hours, there are 2,652 cells present in the culture.

Next, we need to find the derivative p'(t) to understand the rate of change of the cell population. Using the power rule, we differentiate:

p'(t) = 0 + 4t

Now, we evaluate the derivative at t = 24:

p'(24) = 4(24) = 96

This means that after 24 hours, the rate of change of the number of cells is 96 cells per hour. In other words, the cell population is increasing at a rate of 96 cells for every hour that passes.

Understanding these calculations helps in interpreting the growth dynamics of cell cultures, emphasizing the importance of derivatives in analyzing real-world phenomena.

In this scenario, we are analyzing a business's total sales over time, represented by the function \( s(t) = 2.4t^3 + 0.4t^2 + 5t - 1 \), where \( s(t) \) is measured in thousands of dollars and \( t \) is the number of months from now. To understand the sales after 6 months, we need to evaluate \( s(6) \).

Calculating \( s(6) \) involves substituting \( t \) with 6 in the sales function:

\[s(6) = 2.4(6^3) + 0.4(6^2) + 5(6) - 1\]

After performing the calculations, we find:

\[s(6) = 2.4(216) + 0.4(36) + 30 - 1 = 561.8\]

This result indicates that the total sales after 6 months amount to \( 561.8 \) thousand dollars, or \( 561,800 \) dollars. This figure provides insight into the business's performance during this period.

Next, we need to determine the derivative of the sales function, \( s'(t) \), to analyze the rate of change of sales. Using the power rule for differentiation, we derive:

\[s'(t) = 7.2t^2 + 0.8t + 5\]

Now, we evaluate the derivative at \( t = 6 \) to find \( s'(6) \):

\[s'(6) = 7.2(6^2) + 0.8(6) + 5\]

Calculating this gives:

\[s'(6) = 7.2(36) + 4.8 + 5 = 269\]

This result indicates that the rate of change of sales after 6 months is \( 269 \) thousand dollars per month, or \( 269,000 \) dollars per month. This information is crucial as it reflects the business's growth trajectory, suggesting that sales are increasing at a significant rate at this point in time.