Derivative Examples: A Step-by-Step Guide

Hey guys! Ever wondered how to find the derivative of a function? Derivatives are super important in calculus and have tons of applications in physics, engineering, economics, and more. In simple terms, the derivative of a function tells you how much the function's output changes when you make a tiny change to its input. Think of it like the slope of a curve at a specific point. In this guide, we'll walk through a bunch of examples to help you master the art of finding derivatives. So, buckle up, and let's dive in!

Basic Power Rule Examples

Let's kick things off with the power rule, one of the most fundamental rules in calculus. The power rule states that if you have a function in the form f(x) = x^n, where n is any real number, then the derivative f'(x) is given by nx^(n-1). Basically, you multiply by the exponent and then reduce the exponent by one. Sounds simple, right? Let’s see it in action.

Example 1: f(x) = x^3

Here, f(x) = x^3. According to the power rule, we multiply by the exponent (which is 3) and reduce the exponent by one: f'(x) = 3x^(3-1) = 3x^2. Easy peasy! This means that the rate of change of the function f(x) = x^3 at any point x is given by 3x^2. For instance, at x = 2, the rate of change is 3(2^2) = 12.

Example 2: g(x) = x^7

Next up, we have g(x) = x^7. Applying the power rule again, we get g'(x) = 7x^(7-1) = 7x^6. Notice the pattern? The exponent comes down as a coefficient, and the new exponent is one less than the original. This is the core of the power rule. What this tells us is that the function g(x) = x^7 changes at a rate defined by 7x^6. So, if you were to plot this function, the slope at any given point x would be 7x^6.

Example 3: h(x) = x^(-2)

Now, let’s try one with a negative exponent: h(x) = x^(-2). Applying the power rule, we get h'(x) = -2x^(-2-1) = -2x^(-3). We can rewrite this as h'(x) = -2/x^3. Don’t let the negative exponent scare you; the rule still applies! The function h(x) = x^(-2), which can also be written as 1/x^2, has a rate of change given by -2/x^3. As x becomes very large, the rate of change approaches zero.

Example 4: p(x) = √x

What about square roots? Remember that √x can be written as x^(1/2). So, p(x) = x^(1/2). Applying the power rule, we get p'(x) = (1/2)x^((1/2)-1) = (1/2)x^(-1/2). This can be rewritten as p'(x) = 1/(2√x). So, the derivative of the square root function is 1/(2√x). This is a handy one to remember!

Example 5: q(x) = 1/x

Finally, consider q(x) = 1/x, which can be written as x^(-1). Applying the power rule, we have q'(x) = -1x^(-1-1) = -1x^(-2). This simplifies to q'(x) = -1/x^2. The rate of change of the reciprocal function 1/x is -1/x^2. As x increases, the rate of change becomes less negative, approaching zero.

Constant Multiple Rule Examples

The constant multiple rule states that if you have a function in the form f(x) = c g(x), where c is a constant, then the derivative f'(x) = c g'(x). In other words, you can pull the constant out of the derivative. This rule is super useful when dealing with functions that have a constant coefficient.

Example 1: f(x) = 5x^2

Let’s start with f(x) = 5x^2. Here, c = 5 and g(x) = x^2. First, find the derivative of g(x), which is g'(x) = 2x. Now, multiply by the constant: f'(x) = 5 * 2x = 10x. So, the derivative of 5x^2 is 10x. This means that the rate of change of this function is directly proportional to x, with a factor of 10.

Example 2: g(x) = -3x^4

Next, let’s look at g(x) = -3x^4. Here, c = -3 and g(x) = x^4. The derivative of g(x) is g'(x) = 4x^3. Multiply by the constant: g'(x) = -3 * 4x^3 = -12x^3. Therefore, the derivative of -3x^4 is -12x^3. Notice how the negative sign is preserved throughout the process.

Example 3: h(x) = (1/2)x^6

Now, consider h(x) = (1/2)x^6. Here, c = 1/2 and g(x) = x^6. The derivative of g(x) is g'(x) = 6x^5. Multiply by the constant: h'(x) = (1/2) * 6x^5 = 3x^5. So, the derivative of (1/2)x^6 is 3x^5. The constant multiple rule makes it straightforward to deal with fractional coefficients.

Example 4: p(x) = 7√x

Let's try p(x) = 7√x. Rewriting this, we get p(x) = 7x^(1/2). Here, c = 7 and g(x) = x^(1/2). The derivative of g(x) is g'(x) = (1/2)x^(-1/2) = 1/(2√x). Multiply by the constant: p'(x) = 7 * (1/(2√x)) = 7/(2√x). Thus, the derivative of 7√x is 7/(2√x).

Example 5: q(x) = -4/x

Finally, let’s take q(x) = -4/x. Rewriting, we have q(x) = -4x^(-1). Here, c = -4 and g(x) = x^(-1). The derivative of g(x) is g'(x) = -1x^(-2) = -1/x^2. Multiply by the constant: q'(x) = -4 * (-1/x^2) = 4/x^2. The derivative of -4/x is 4/x^2. The negative signs canceled out in this case.

Sum and Difference Rule Examples

The sum and difference rule is used when you have a function that is the sum or difference of two or more functions. The rule states that if f(x) = u(x) + v(x), then f'(x) = u'(x) + v'(x), and if f(x) = u(x) - v(x), then f'(x) = u'(x) - v'(x). In simpler terms, you can take the derivative of each term separately and then add or subtract them.

Example 1: f(x) = x^3 + 2x^2

Let's start with f(x) = x^3 + 2x^2. Here, u(x) = x^3 and v(x) = 2x^2. The derivative of u(x) is u'(x) = 3x^2, and the derivative of v(x) is v'(x) = 4x. Therefore, f'(x) = 3x^2 + 4x. This means that the rate of change of the function f(x) is the sum of the rates of change of its individual terms.

Example 2: g(x) = 4x^5 - 3x^2 + 2x - 1

Next, consider g(x) = 4x^5 - 3x^2 + 2x - 1. We can break this into multiple terms: u(x) = 4x^5, v(x) = -3x^2, w(x) = 2x, and z(x) = -1. The derivatives are u'(x) = 20x^4, v'(x) = -6x, w'(x) = 2, and z'(x) = 0. So, g'(x) = 20x^4 - 6x + 2. Notice how the constant term (-1) disappears upon differentiation.

Example 3: h(x) = √x - 1/x

Let’s look at h(x) = √x - 1/x. Rewriting this, we have h(x) = x^(1/2) - x^(-1). The derivatives are (1/2)x^(-1/2) and -(-1)x^(-2) = x^(-2). Thus, h'(x) = (1/2)x^(-1/2) + x^(-2), which can be written as h'(x) = 1/(2√x) + 1/x^2. The sum and difference rule allows us to handle each term separately, making the process more manageable.

Example 4: p(x) = 5x^3 + 2√x - 3/x^2

Now, let's try p(x) = 5x^3 + 2√x - 3/x^2. Rewriting, we get p(x) = 5x^3 + 2x^(1/2) - 3x^(-2). The derivatives are 15x^2, 2 * (1/2)x^(-1/2) = x^(-1/2), and -3 * (-2)x^(-3) = 6x^(-3). So, p'(x) = 15x^2 + x^(-1/2) + 6x^(-3), which can be written as p'(x) = 15x^2 + 1/√x + 6/x^3. Each term's derivative is added or subtracted according to the original function.

Example 5: q(x) = x^4 - 6x^(1/3) + 10

Finally, let’s take q(x) = x^4 - 6x^(1/3) + 10. The derivatives are 4x^3, -6 * (1/3)x^(-2/3) = -2x^(-2/3), and 0. So, q'(x) = 4x^3 - 2x^(-2/3), which can be written as q'(x) = 4x^3 - 2/(x^(2/3)). Constant terms vanish, and each power term is handled using the power rule.

Conclusion

So, there you have it! We've covered the basic power rule, the constant multiple rule, and the sum and difference rule with plenty of examples. These rules form the foundation for finding derivatives of many common functions. Keep practicing, and you'll become a derivative pro in no time! Remember, understanding these rules and how to apply them is key to mastering calculus. Happy differentiating!