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Derivatives

- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
- Accumulation Function
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- Algebraic Functions
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- Antiderivatives
- Application of Derivatives
- Approximating Areas
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- Derivatives
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- Determining Volumes by Slicing
- Direction Fields
- Disk Method
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- Eliminating the Parameter
- Euler's Method
- Evaluating a Definite Integral
- Evaluation Theorem
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- First Derivative Test
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- Expression Math
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- Faces Edges and Vertices
- Factorials
- Factoring Polynomials
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- Factorising expressions
- Factors
- Finding Maxima and Minima Using Derivatives
- Finding Rational Zeros
- Finding the Area
- Forms of Quadratic Functions
- Fractional Powers
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- Fractions and Factors
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- Function Basics
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- Iterative Methods
- Law of Cosines in Algebra
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One of the fastest cars in the world, Hennessey Performance Engineering's Venom F5 runs at a top speed of greater than \( 300 mph \) and can go from \(0\) to \( 62 mph \) in just \( 2.6 \) seconds!^{1}

In this article, you will explore the techniques that can be used to calculate the acceleration the Venom F5 achieves while approaching its top speed.

- The purpose of derivatives.
- Defining the derivative.
- The derivative formula.
- The derivative rules.
- Examples of derivatives.

Using its most basic definition, the purpose of a derivative is to **calculate the slope of a tangent line to a curve**.

But what does this mean, exactly?

It means that you are calculating the **rate of change** of a function at a point. More specifically, it could mean calculating velocity and changes in velocity over time, like you would want to do for the Venom F5. However, the uses of derivatives are far more varied:

They are useful in all branches of

**mathematics**,**science**, and**engineering**.Derivatives are crucial in

**business****analyses**.They are vital in

**health**applications.

Now that you have a good grasp on Limits, you have established the foundation for your study of Calculus!

To truly begin the study of calculus, you must revisit the concepts of **secant **and **tangent ****lines** to a curve.

You can calculate the slope of a secant line to a curve in one of two ways, as shown by the graphs and formulas below. Since both methods give the same answer, you can choose whichever method is easiest to calculate.

These formulas for the slope of the secant line to a curve are in the form of a **difference quotient**. Their formal definition is:

- Let \( f \) be a function that is defined on the interval \( I \), and let \( I \) contain the value \( a \). If \( x \) is in \( I \) and \( x \neq a \) , then the
**difference quotient**is:\[ Q = \frac{f(x)-f(a)}{x-a} \] - And, the
**difference quotient****with an increment of \( \bf{ h } \)**, that exists if \( h \neq 0 \), is chosen such that \( a+h \) is in the interval \( I \), is:\[ Q = \frac{f(a+h)-f(a)}{h} \]

As you move the second point – \( (x, f(x)) \) in the top graph above, and \( (a+h, f(a+h)) \) in the bottom graph above – of the secant line closer to the first point – \( (a, f(a)) \) in both graphs above – you **approach **the **tangent ****line **to the curve at that first point. This leads you to use **limits **to define the tangent line to a curve as:

- Let \( f(x) \) be a function that is defined on an open interval that contains \( a \). The
**tangent****line**to \( f(x) \) at \( a \) is the line that passes through the point \( (a, f(a)) \) and has a slope of:\[ m_{tan} = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \]if the limit exists. - And, if the limit exists, you can also define the
**tangent****line**to \( f(x) \) at \( a \) as the line that passes through the point \( (a, f(a)) \) and has a slope of:\[ m_{tan} = \lim_{h \to 0} \frac{f(a+h)}{h}. \]

For a more in-depth review and examples, please refer to the article on Tangent Lines.

This use of limits to find the slope of the tangent line to a function (or curve) at a point brings you right to the **derivative**. Essentially, the term derivative is just a special name for this type of limit, and the process of finding (also called taking) a derivative is called **differentiation**. The **definition of the derivative of a function at a point** is:

- The
**derivative of a function**\( f(x) \)**at the point**\( a \), which is denoted by \( f'(a) \) (read as “f-prime of a”), and provided that \( f(x) \) is a function defined on an open interval which contains \( a \), and that the limit exists, is defined as:\[ f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}. \] - And, again if the limit exists, you can also define the derivative of \( f(x) \) at \( a \) as:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}. \]

Just as with secant and tangent lines to the curve, you can choose to use either of these formulas to find the derivative of a function at a point.

Find \( f'(2) \) for the function:

\[ f(x) = 3x^{2}-4x+1. \]

**Solutions**:

- Using the formula:\[ f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \]
- Apply the definition (here, \( a = 2 \)).\[ f'(2) = \lim_{x \to 2} \frac{f(x)-f(2)}{x-2} \]
- Substitute \( f(x) = 3x^{2}-4x+1 \) and \( f(2) = 5 \).\[ f'(2) = \lim_{x \to 2} \frac{(3x^{2}-4x+1)-5}{x-2} \]
- Simplify the numerator.\[ f'(2) = \lim_{x \to 2} \frac{3x^{2}-4x-4}{x-2} \]
- Factor the numerator.\[ f'(2) = \lim_{x \to 2} \frac{(x-2)(3x+2)}{x-2} \]
- Cancel the common factor.\[ f'(2) = \lim_{x \to 2} (3x+2) \]
- Substitute \( x = 2 \).\[ f'(2) = \lim_{x \to 2} (3(2)+2) \]
- Evaluate the limit.\[ \bf{ f'(2) } = \bf{ 8 } \]

- Using the formula:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h} \]
- Apply the definition (here, \( a = 2 \)).\[ f'(2) = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} \]
- Substitute \( f(2+h) = 3(2+h)^{2}-4(2+h)+1 \) and \( f(2) = 5 \).\[ f'(2) = \lim_{h \to 0} \frac{(3(2+h)^{2}-4(2+h)+1)-5}{h} \]
- Simplify the numerator.\[ f'(2) = \lim_{h \to 0} \frac{3h^{2}+8h}{h} \]
- Factor the numerator.\[ f'(2) = \lim_{h \to 0} \frac{h(3h+8)}{h} \]
- Cancel the common factor.\[ f'(2) = \lim_{h \to 0} (3h+8) \]
- Substitute \( h = 0 \).\[ f'(2) = \lim_{h \to 0} (3(0)+8) \]
- Evaluate the limit.\[ \bf{ f'(2) } = \bf{ 8 } \]

Notice that using either method you get exactly the same answer!

One of the most common ways you can use derivatives is to find **velocities **and **rates of change**. If you have a function that represents an object's position over time, called \( s(t) \), then it's average velocity is given by the difference quotient:

\[ v_{avg} = \frac{s(t)-s(a)}{t-a}. \]

As the values of \(t\) approach the value of \(a\), the value of \(v_{avg}\) approaches what you call the **instantaneous velocity** at \(a\). The instantaneous velocity at \(a\), denoted by \(v(a)\), is defined as:

\[ v(a) = s'(a) = \lim_{t \to a} \frac{s(t)-s(a)}{t-a} .\]

In other words, if you want to find the average velocity of an object over time, you can do so by taking the derivative of the function that represents the object's position over time.

The graph above helps you understand the **difference between average velocity and instantaneous velocity**. In this graph, the **slope of the secant line** (green) is the **average velocity** of the object over the interval of time \([a, t]\). The **slope of the tangent line **(pink) is the **instantaneous velocity** of the object at time \( t = a \). The position of the object at time \(t\) is represented by the function \(y = s(t)\) (blue).

As you can see, the slope of the tangent line to a function and instantaneous velocity are related. You calculate both by finding a derivative, and both measure the **instantaneous rate of change** of a function – the rate of change of said function at a specific point.

The **instantaneous rate of change** of a function, \( f(x) \), at a value of \(a\) is its **derivative ****at \(a\)**. In other words the instantaneous rate of change of \(f(x)\) at \(a\) has the value \(f'(a)\).

Using the formula for \(v(a)\), you can calculate an object's instantaneous velocity, or you can use a table of values to estimate the velocity of a moving object. You can confirm this estimate using the formula for \(v_{avg}\).

Say you have a weight on a spring that is oscillating up and down. Its position at time \(t\) with respect to a fixed horizontal line (or datum) is given by the function:

\[ s(t) = sin(t). \]

Using a table of values, estimate \(v(0)\). Check this estimate using the formula for \(v(a)\).

**Solution**:

You can estimate the instantaneous velocity at time \(t=0\) by creating a table of values, choosing values of \(t\) that approach \(0\), like in the table below.

\(t\) | \( \frac{sin(t)-sin(0)}{t-0} = \frac{sin(t)}{t} \) |

\(-0.1\) | \(0.998334166\) |

\(-0.01\) | \(0.999983333\) |

\(-0.001\) | \(0.999999833\) |

\(0.001\) | \(0.999999833\) |

\(0.01\) | \(0.999983333\) |

\(0.1\) | \(0.998334166\) |

From this table, you can see that the average velocity over intervals of time that approach \(t=0\) approaches a velocity of \(1\), so it seems that \(v(0)=1\) is a good estimate.

You can check this estimate using the formula:

\[ v(a) = s'(a) = \lim_{t \to a} \frac{s(t)-s(a)}{t-a}. \]

Using this formula, you get:

\[ \begin{align}v(0) = s'(0) &= \lim_{t \to 0} \frac{sin(t)-sin(0)}{t-0} \\&= \lim_{t \to 0} \frac{sin(t)}{t} \\&= 1.\end{align} \]

Therefore, \( \bf{ v(0) } = \bf{ 1 } \).

So far, you have looked at the derivative of a function at a specific point. By considering the derivative itself as a function, you can find the derivative of a function at every point in its domain for which the derivative is defined (the limit exists). Below, you formally define a **derivative function**.

There is a formula you can use to find any derivative of any function. It is defined as:

Let \(f(x)\) be a function. The derivative of this function with respect to \(x\), denoted by \(f'(x)\), is the function whose domain consists of values of \(x\) such that the following limit exists:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}. \]

This is called the **definition of the derivative**.

Of course, as you progress in your study of calculus, you will discover more efficient ways to calculate the derivatives of functions that take on certain forms. Those are touched on these a little later in this article, and more in-depth explanations and analyses are in the article on Differentiation Rules and its sub-articles.

Using the formula for the definition of a derivative will always work, though it might take longer to work through finding the derivative this way!

Using the definition of a derivative to find the derivative function of:

\[ f(x) = x^{2}-2x. \]

**Solution**:

- Substitute \( f(x+h) = (x+h)^{2}-2(x+h) \) and \( f(x) = x^{2}-2x\) into the definition of the derivative.\[ f'(x) = \lim_{h \to 0} \frac{((x+h)^{2}-2(x+h))-(x^{2}-2x)}{h} \]
- Expand \( (x+h)^{2}-2(x+h)\).\[ f'(x) = \lim_{h \to 0} \frac{x^{2}+2xh+h^{2}-2x-2h-x^{2}+2x}{h} \]
- Simplify the numerator.\[ f'(x) = \lim_{h \to 0} \frac{2xh-2h+h^{2}}{h} \]
- Factor an \(h\) out of the numerator.\[ f'(x) = \lim_{h \to 0} \frac{h(2x-2+h)}{h} \]
- Cancel the common factor of \(h\).\[ f'(x) = \lim_{h \to 0} (2x-2+h) \]
- Evaluate the limit by using direct substitution.\[ \bf{ f'(x) } = \bf{ 2x-2 } \]

For an in-depth analysis and more examples, please refer to the article on Derivative Functions.

Since a function and its derivative are related, you would expect that their graphs are related as well, because the derivative gives us the rate of change of the function. This is best seen with an example.

Graph the function \( f(x) = x^{2}-2x \) and its derivative \( f'(x) = 2x-2 \).

**Solution**:

With the function (blue) and its derivative (green) graphed on the same coordinate plane, you can see their relationship.

- You can see that \( f(x) \) is decreasing for values of \( x < 1 \).
- Therefore, for those same values of \(x\), \( f'(x) < 0 \).

- You can also see that \( f(x) \) is increasing for values of \( x > 1 \).
- Therefore, for those same values of \(x\), \( f'(x) > 0 \).

- And, \( f(x) \) has a horizontal tangent line at \( x=1 \).
- Here, \( f'(1)=0 \).

Now that you have graphed a derivative, let's discuss the behavior of these graphs. The first things to consider are **differentiability **and **continuity**. Let's start with the theorem:

Let \( f(x) \) be a function that has \(a\) in its domain. If \(f(x)\) is differentiable at \(a\), then it is also continuous at \(a\).

Note that this theorem DOES NOT work in reverse! Differentiability implies continuity, but continuity does not imply differentiability.

Proof of the theorem: Differentiability Implies Continuity.

If \(f(x)\) is differentiable at \(a\), then \(f'(a)\) exists and:

\[ f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}. \]

To prove this, we need to show that \(f(x)\) is continuous at \(a\) by showing that \( \lim_{x \to a} (f(x)) = f(a) \). So,

\[ \begin{align}\lim_{x to a} (f(x)) &= \lim_{x \to a}(f(x)-f(a)+f(a)) \\&= \lim_{x \to a} \left( \frac{f(x)-f(a)}{x-a} \times (x-a) + f(a) \right) \\&= \left( \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \right) \times \left( \lim_{x \to a} (x-a) \right) + \lim_{x \to a} f(a) \\&= f'(a) \times 0 + f(a) \\&= f(a).\end{align} \]

Since \( f(a) \) is defined and \( \lim_{x \to a} (f(x)) = f(a) \), you can conclude that \( f(x) \) is continuous at \(a\).

Continuity does not imply differentiability.

Let's look at the absolute value function:

\[ f(x) = |x|. \]

You know this function is continuous everywhere, but is it differentiable everywhere? Let's try to take the derivative at the point where \( x=0 \).

\[ \begin{align}f'(0) &= \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} \\&= \lim_{x \to 0} \frac{|x|-|0|}{x-0} \\&= \lim_{x \to 0} \frac{|x|}{x}.\end{align} \]

This limit does not exist because

\[ \lim_{x \to 0^{-}} \frac{|x|}{x} = -1 \]

and

\[ \lim_{x \to 0^{+}} \frac{|x|}{x} = 1. \]

In other words, as you approach the limit from the left-hand side of the graph, the limit is \(-1\), and as you approach the limit from the right-hand side of the graph, the limit is \(+1\). In order for the limit to exist, it needs to be consistent no matter from which side you approach the limit.

This was just one example where continuity does not imply differentiability. A summary of situations where a continuous function is not differentiable include:

- As with the absolute value function, if the limit of the slopes of the tangent lines to the curve on the left and right are not the same, the function is not differentiable.
- In the case of the absolute value function, this resulted in a sharp corner of the graph at \(0\). This leads us to conclude that for a function to be differentiable at a point, it must be “smooth” at that point.

- A function is not differentiable at any point where it has a tangent line that is vertical.
- A function may fail to be differentiable in more complicated ways, such as a function whose oscillations become increasingly frequent as it approaches a value.

Since the derivative itself is a function, it is possible to find the derivative of a derivative. The most common example of this is when talking about **position**, **velocity**, and **acceleration**:

The derivative of a position function is the rate of change of position, or velocity.

The derivative of a velocity function is the rate of change of velocity, or acceleration.

When you take the derivative of a derivative, the new function is called the **second derivative**. You can continue this process to find the **third derivative**, the **fourth derivative**, etc. Those types of derivatives are referred to collectively as **Higher-Order Derivatives**. The notation of higher-order derivatives of the function \( y = f(x) \) can be expressed in any of the following formats:

\[ f''(x), f'''(x), f^{(4)}(x), \cdots, f^{(n)}(x) \]

\[ y''(x), y'''(x), y^{(4)}(x), \cdots, y^{(n)}(x) \]

\[ \frac{d^2y}{dx^2}, \frac{d^3y}{dx^3}, \frac{d^4y}{dx^4}, \cdots, \frac{d^ny}{dx^n} \]

For an in-depth analysis and examples, please refer to the article on Higher-Order Derivatives.

As mentioned earlier, it is possible to find the derivatives of functions using just the definition of the derivative. That process, however, can be quite lengthy and sometimes quite challenging! This section introduces you to the rules for finding derivatives without having to use the definition of the derivative every time.

The basic functions \( f(x)=c \) and \( f(x)=x^n \), where \(n\) is a positive integer, are the building blocks from which every polynomial and rational functions are built. If you want to find their derivatives without using the definition of the derivative, you need to first develop formulas for differentiating these functions.

Let's start with the most basic function, the constant function: \( f(x)=c \). The rule for differentiating constant functions is called **the constant rule**.

**The constant rule states that the derivative of any constant function is zero**. That is, since a constant function is a horizontal line, the slope (or rate of change) of a constant function is zero.

**Theorem – The Constant Rule**

Let \(c\) be any constant. If \( f(x)=c \), then the derivative of \(f(x)\) can be written in any of the following ways:

\[ f'(c)=0 \]

\[ y'(c)=0 \]

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

Just like when you work with functions, there are rules that help you find the derivatives of functions that you add, subtract, or multiply by a constant.

**Theorem – Sum, Difference, and Constant Multiple Rules**

Let \( f(x) \) and \( g(x) \) be differentiable functions, and let \(c\) be a constant. Then, each of the following holds true:

**Sum Rule**– The derivative of the sum of two functions, \( f(x) \) and \( g(x) \), is equal to the sum of the derivatives of \( f(x) \) and \( g(x) \).\[ \frac{d}{dx} (f(x)+g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x)) \] or, if you have\[ h(x) = f(x) + g(x), \]then\[ h'(x) = f'(x) + g'(x). \]**Difference Rule**– The derivative of the difference of two functions, \( f(x) \) and \( g(x) \), is equal to the difference of the derivatives of \( f(x) \) and \( g(x) \).\[ \frac{d}{dx} (f(x)-g(x)) = \frac{d}{dx}(f(x))-\frac{d}{dx}(g(x)) \] or, if you have\[ h(x) = f(x) - g(x) \]then\[ h'(x) = f'(x) - g'(x). \]**Constant Multiple Rule**– The derivative of a constant \(c\) multiplied by a function is equal to the constant multiplied by the derivative of the function.\[ \frac{d}{dx}(c \cdot f(x)) = c \cdot \frac{d}{dx}(f(x)) \] or, if you have\[ h(x) = c \cdot f(x) \]then\[ h'(x) = c \cdot f'(x). \]

For an in-depth analysis and examples of these basic derivative rules, please refer to the article on Differentiation Rules.

The next basic function is \( f(x)=x^n \). There are several ways you can find the power rule provided you restrict \(n\) to a positive integer, however, to prove the power rule for any real number \(n\), you need to understand both Logarithmic Differentiation and Implicit Differentiation.

**Theorem – The Power Rule**

Let \(n\) be any constant. If \( f(x)=x^n \), then the derivative of \( f(x) \) can be written in any of the following ways:

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

\[ y'(x)=nx^{n-1} \]

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

For an in-depth analysis and examples, please refer to the article on The Power Rule.

The product rule states that the derivative of the product of two functions is:

The derivative of the first function times the second function

*plus*the first function times the derivative of the second function.

**Theorem – The Product Rule**

Let \( f(x) \) and \( g(x) \) be differentiable functions. Then,

\[ \frac{d}{dx}(f(x)g(x)) = \frac{d}{dx}(f(x))g(x) + f(x) \frac{d}{dx}(g(x)), \] or, if you have

\[ h(x) = f(x)g(x) \]

then

\[ h'(x) = f'(x)g(x) + f(x)g'(x). \]

While it may be tempting to think so, the product rule DOES NOT follow the pattern of the sum/difference rules. The derivative of a product IS NOT the product of the derivatives!

For an in-depth analysis and examples, please refer to the article on The Product Rule.

Now that you have looked at the derivatives of the product of two functions, let's consider the quotients of functions, and develop The Quotient Rule.

The quotient rule states that the derivative of the quotient of two functions is:

The derivative of the function in the numerator times the function in the denominator

*minus*the function in the numerator times the derivative of the function in the denominator,All divided by the function in the denominator squared.

**Theorem – The Quotient Rule**

Let \( f(x) \) and \( g(x) \) be differentiable functions. Then,

\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{\frac{d}{dx}(f(x))g(x)-f(x)\frac{d}{dx}(g(x))}{(g(x))^2} \] or, if you have

\[ h(x) = \frac{f(x)}{g(x)} \]

then

\[ h'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}. \]

While it may be tempting to think so, the quotient rule DOES NOT follow the pattern of the sum/difference rules. The derivative of a quotient IS NOT the quotient of the derivatives!

For an in-depth analysis and examples, please refer to the article on The Quotient Rule.

As you continue your study of calculus, you will see that it is rare to use only one differentiation rule to find the derivative of a given function. Using the rules you have learned, you can now find the derivatives of polynomial and rational functions.

A general rule of thumb to use when applying several differentiation rules is to apply them in reverse of the order in which we would want to evaluate the function.

If \( f(x) = 3g(x)+x^{2}h(x) \), find \( f'(x) \).

**Solution**:

- Set up the problem.\[ f'(x) = \frac{d}{dx} \left( 3g(x) + x^{2}h(x) \right) \]
- Apply the sum rule.\[ f'(x) = \frac{d}{dx} (3g(x)) + \frac{d}{dx}(x^{2}h(x)) \]
- Apply the constant multiple rule and the product rule.\[ f'(x) = 3 \frac{d}{dx}(g(x)) + \left( \frac{d}{dx}(x^{2})h(x) + x^{2} \frac{d}{dx}(h(x)) \right) \]
- Take the derivatives.\[ f'(x) = 3g'(x) + 2xh(x) + x^{2}h'(x) \]

For an in-depth analysis and more examples, please refer to the article on Combining Differentiation Rules.

Let's take a look at some examples when using derivatives.

First, one involving an instantaneous rate of change.

Say you set the thermostat of your home so that the temperature drops from \(70^{\circ} F\) at \(9pm\), reaches a low of \(60^{\circ} F\) during the night, and then rises back up to \(70^{\circ} F\) by \(7am\) the next morning. Now suppose that the equation:

\[ T(t) = 0.4t^{2}-4t+70 \]

for \( 0 \leq t \leq 10 \)

where \( t = \) number of hours past \( 9pm \) represents the temperature of your home throughout the night.

Find the instantaneous rate of change of the temperature at midnight.

**Solution**:

Since midnight is \(3\) hours after \(9pm\), you want to find the instantaneous rate of change, or the derivative, of the equation at \(t=3\). In other words, you need to calculate \(T'(3)\).

\[ \begin{align}T'(3) &= \lim_{t \to 3} \left( \frac{T(t)-T(3)}{t-3} \right) \\&= \lim_{t \to 3} \left( \frac{0.4t^{2}-4t+70-61.6}{t-3} \right) \\&= \lim_{t \to 3} \left( \frac{0.4t^{2}-4t+8.4}{t-3} \right) \\&= \lim_{t \to 3} \left( \frac{0.4(t-3)(t-7)}{t-3} \right) \\&= \lim_{t \to 3} (0.4(t-7)) \\&= (0.4(3-7)) \\T'(3) &= -1.6\end{align} \]

Therefore, the **instantaneous rate of change of the temperature at midnight is \( -1.6^{\circ} F \) per hour**.

How do you find the derivative of the square root function?

Find the derivative of the square-root function \( f(x)=\sqrt{x} \).

**Solution**:

- Apply the definition of the derivative.\[ f'(x) = \lim_{h \to 0} \left( \frac{f(x+h)-f(x)}{h} \right) \]
- Substitute \( f(x)=\sqrt{x} \).\[ f'(x) = \lim_{h \to 0} \left( \frac{\sqrt{x+h}-\sqrt{x}}{h} \right) \]
- Multiply and divide by \( \sqrt{x+h}+\sqrt{x} \) so you can simplify the fraction.\[ f'(x) = \lim_{h \to 0} \left( \frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \right) \]
- Simplify.\[ f'(x) = \lim_{h \to 0} \left( \frac{h}{h \left( \sqrt{x+h} + \sqrt{x} \right)} \right) \]
- Cancel out the common factor of \(h\).\[ f'(x) = \lim_{h \to 0} \left( \frac{1}{\sqrt{x+h}+\sqrt{x}} \right) \]
- Evaluate the limit.\[ f'(x) = \frac{1}{\sqrt{x}+\sqrt{x}} \]
- Simplify.\[ \bf{ f'(x) } = \bf{ \frac{1}{2\sqrt{x}} } \]

Now let's see how the square root function and its derivative compare graphically.

Graph the function \( f(x)=\sqrt{x} \) and its derivative \( f'(x) = \frac{1}{2\sqrt{x}}.

**Solution**:

With the function (blue) and its derivative (green) graphed on the same coordinate plane, you can see their relationship.

- First, notice that \( f(x) \) is increasing over its entire domain. This means the slopes of all the tangent lines are positive.
- Therefore, you expect \( f'(x) > 0 \) for all values of \(x\) in its domain.

- Second, as \(x\) increases, the slopes of the tangent lines to the curve of \(f(x)\) are decreasing.
- Therefore, you expect \(f'(x)\) to decrease as well.

- Finally, you can see that \(f(0)\) is undefined and that \( \lim_{x \to 0^{+}} f'(x) = \pm \infty \).
- This corresponds to the vertical tangent to \(f(x)\) at \(0\).

Finally, let's work through a position, velocity, acceleration example involving higher-order derivatives.

The position of an object along a coordinate axis at time \(t\) (in seconds) is given by the function \( s(t) = 3t^{2}-4t+1 \) (in meters). Find the function that describes its acceleration at time \(t\).

**Solution**:

To find the acceleration function from a position function, you need to take the first and second derivatives of the position function.

- The first derivative of the position function is its velocity function: \( s'(t) = v(t) \).\[ \begin{align}s'(t) &= \lim_{h \to 0} \left( \frac{s(t+h)-s(t)}{h} \right) \\&= \lim_{h \to 0} \left( \frac{3(t+h)^{2}-4(t+h)+1-(3t^{2}-4t+1)}{h} \right) \\&= \lim_{h \to 0} \left( \frac{3t^{2}+6th+3h^{2}-4t-4h+1-3t^{2}+4t-1}{h} \right) \\&= \lim_{h \to 0} \left( \frac{h(6t+3h-4)}{h} \right) \\&= \lim_{h \to 0} (6t+3h-4) \\s'(t) &= 6t-4 \\\end{align} \]
- Therefore, \( v(t) = 6t-4 \).

- The second derivative of the position function is its acceleration function: \( s''(t) = v'(t) = a(t) \).\[ \begin{align}s''(t) &= \lim_{h \to 0} \left( \frac{s'(t+h)-s'(t)}{h} \right) \\&= \lim_{h \to 0} \left( \frac{6(t+h)-4-(6t-4)}{h} \right) \\&= \lim_{h \to 0} \left( \frac{6t+6h-4-6t+4}{h} \right) \\&= \lim_{h \to 0} \left( \frac{6h}{h} \right) \\&= \lim_{h \to 0} (6) \\s''(t) &= 6\end{align} \]
- Therefore, \( \bf{ a(t) } = \bf{ 6 \frac{m}{s^{2}} } \).

- The purpose of a derivative is to
**calculate the slope of a tangent line to a curve at a specific point**. - The
**instantaneous rate of change**of a function, \( f(x) \), at a value of \(a\) is its**derivative****at \(a\)****, \( f'(a) \)**. - The definition of a derivative is:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \]
- Differentiability implies continuity, but continuity does not imply differentiability.
- You can take the derivative of a derivative
- This results in Higher-Order Derivatives

- There exist several derivative rules to help us find the derivatives of functions more quickly and easily.

- https://www.hennesseyspecialvehicles.com/2021-venom-f5/

In mathematics, a derivative is the limit we calculate to find the slope of the tangent line to a function at a specific point.

Put simply, a derivative is the calculation of the rate of change of a function.

The formula for a derivative is:

f'(a) = lim(h⇾0) {[f(a+h)-f(a)]/h}

There are 2 ways to find the derivative of a function:

- Using the definition of a derivative
- Make sure the function is in its simplest form. This makes it easier to work with!
- Apply the formula for the definition of a derivative.

- Using the differentiation rules
- Make sure the function is in its simplest form. This makes it easier to work with!
- Identify the form of the function you need to differentiate.
- A number (e.g., 5)
- A number multiplied by a variable (e.g., 5x)
- A number multiplied by a variable with an exponent (e.g., 5x^2)
- Addition (e.g., 5x + 5)
- Multiplication of variables (e.g., the form x·x)
- Division of variables (e.g., the form x/x)

- Apply the differentiation rule that makes the most sense for the form of your function.

- Choose a point on the graph at which you want to find the derivative.
- Draw a straight line that is tangent to the curve of the graph at this point.
- Calculate the slope of the straight line you just drew.
- The slope you calculated is the derivative of the graph at this point!

The derivative of a constant is the constant itself.

More about Derivatives

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