Derivative Rules of Basic Functions. Visual Explanation with color coded examples (2024)

Quick Overview

  • The derivative rules are established using the definition.
  • Once we've established the derivative rules, we can use them to find derivatives more efficiently than using the definition each time.
  • The derivative rules include...
    • Derivative of Constants
    • Derivative of the Identity Function
    • Sum Rule
    • Difference Rule
    • Constant Coefficient Rule
    • Derivatives of Linear Functions
    • Derivatives of Sines, Cosines and Exponential

Derivatives of Constants

Find $$\displaystyle \frac d {dx} \left(k\right)$$

Step 1

Evaluate the definition of the derivative.

$$ \\ \begin{align*} \frac d {dx}\left(k\right) & = \lim_{h\to0} \frac{\blue{f(x+h)} - \red{f(x)}} h\\[6pt] & = \lim_{h\to0} \frac{\blue{k} - \red{k}} h\\[6pt] & = \lim_{h\to0} \frac 0 h\\[6pt] & = \lim_{h\to0} 0\\[6pt] & = 0 \end{align*} \\ $$

Answer

$$\displaystyle \frac d {dx}\left(k\right) = 0$$

Derivative of the Identity Function

Recall that the identity function is $$f(x) = x$$. Find $$f'(x)$$

Step 1

Evaluate the functions in the definition of the derivative. Then simplify.

$$ \begin{align*} f'(x) & = \lim_{h\to 0} \frac{\blue{f(x+h)} -\red{f(x)}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{x+h} -\red x} h\\[6pt] & = \lim_{h\to 0} \frac h h\\[6pt] & = \lim_{h\to 0} 1\\[6pt] & = 1 \end{align*} $$

Answer

$$f'(x) = 1$$ when $$f(x) = x$$

Sum Rule for Derivatives

Suppose $$f(x)$$ and $$g(x)$$ are differentiable1 and $$h(x) = f(x) + g(x)$$. Find $$h'(x)$$.

1If a function is differentiable, then its derivative exists.

Step 1

Evaluate the functions in the definition of the derivative

$$ \begin{align*} h'(x) & = \lim_{\Delta x\to 0} \frac{\blue{h(x+\Delta x)} - \red{h(x)}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x)+g(x+\Delta x)} - \red{[f(x)+g(x)]}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x)+g(x+\Delta x)} - \red{f(x)-g(x)}}{\Delta x} \end{align*} $$

Step 2

Rearrange the numerator so that the $$f$$ terms are together and the $$g$$ terms are together.

$$ \begin{align*} h'(x) & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x)}+\red{g(x+\Delta x)} - \blue{f(x)}-\red{g(x)}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x)} - \blue{f(x)}+\red{g(x+\Delta x)}-\red{g(x)}}{\Delta x} \end{align*} $$

Step 3

Split into two limits. One containing the $$f$$ terms, the other containing the $$g$$ terms.

$$ \begin{align*} h'(x) & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x) - f(x)}+\red{g(x+\Delta x)-g(x)}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \left(\frac{\blue{f(x+\Delta x) - f(x)}}{\blue{\Delta x}}+\frac{\red{g(x+\Delta x)-g(x)}}{\red{\Delta x}}\right)\\[6pt] & = \lim_{\Delta x\to 0}\frac{\blue{f(x+\Delta x) - f(x)}}{\blue{\Delta x}}+ \lim_{\Delta x\to 0}\frac{\red{g(x+\Delta x)-g(x)}}{\red{\Delta x}} \end{align*} $$

Step 4

Evaluate each limit.

Notice that the first limit is exactly the definition of $$f'(x)$$ and the second limit is the definition of $$g'(x)$$

$$ \begin{align*} h'(x) & = \blue{\lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}}+ \red{\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}}\\[6pt] & = \blue{f'(x)}+ \red{g'(x)} \end{align*} $$

Summary of the Rule: $$\displaystyle \frac d{dx}\left( f(x) + g(x)\right) = f'(x) + g'(x)$$

Difference Rule for Derivatives

Suppose $$f(x)$$ and $$g(x)$$ are differentiable and $$h(x) = f(x) - g(x)$$. Find $$h'(x)$$.

Step 1

Evaluate the functions in the definition of the derivative.

$$ \begin{align*} \frac d {dx}\left(f(x) - g(x)\right) & = \lim_{\Delta x\to 0} \frac{\blue{h(x+\Delta x)} - \red{h(x)}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x) - g(x+\Delta x)} - (\red{f(x)-g(x)})}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x) - g(x+\Delta x)} - \red{f(x) + g(x)}}{\Delta x} \end{align*} $$

Step 2

Group the $$f$$ terms and the $$g$$ terms.

$$ \begin{align*} \frac d {dx}\left(f(x) - g(x)\right) & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x)} - \red{g(x+\Delta x)} - \blue{f(x)} + \red{g(x)}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x)} - \blue{f(x)} - \red{g(x+\Delta x)} + \red{g(x)}}{\Delta x} \end{align*} $$

Step 3

Separate into two limits.

$$ \begin{align*} \frac d {dx}\left(f(x) - g(x)\right) & = \lim_{\Delta x\to 0} \frac{\blue{f(x+\Delta x) - f(x)} \red{\,-\,g(x+\Delta x) + g(x)}}{\Delta x}\\[6pt] & = \lim_{\Delta x\to 0} \left(\blue{\frac{f(x+\Delta x) - f(x)}{\Delta x}} + \red{\frac{- g(x+\Delta x) + g(x)}{\Delta x}}\right)\\[6pt] & = \lim_{\Delta x\to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} + \lim_{\Delta x\to 0} \frac{- g(x+\Delta x) + g(x)}{\Delta x}\\[6pt] & = \blue{\lim_{\Delta x\to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}} - \red{\lim_{\Delta x\to 0} \frac{g(x+\Delta x) - g(x)}{\Delta x}} \end{align*} $$

Step 4

Evaluate each limit.

Note that the first limit is precisely the definition of $$f'(x)$$ while the second limit is the definition of $$g'(x)$$.

$$ \begin{align*} \frac d {dx}\left(f(x) - g(x)\right) & = \blue{\lim_{\Delta x\to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}} - \red{\lim_{\Delta x\to 0} \frac{g(x+\Delta x) - g(x)}{\Delta x}}\\[6pt] & = \blue{f'(x)} - \red{g'(x)} \end{align*} $$

Summary of the Rule: $$\displaystyle \frac d {dx}\left(f(x)-g(x)\right) = f'(x) - g'(x)$$

Constant Coefficient Rule

Suppose $$f(x)$$ is differentiable and $$g(x) = k\cdot f(x)$$. Find $$g'(x)$$.

Step 1

Evaluate the functions in the definition of the derivative.

$$ \begin{align*} g'(x) & = \lim_{x\to h} \frac{\blue{g(x+h)} - \red{g(x)}} h\\[6pt] & = \lim_{x\to h} \frac{\blue{k\cdot f(x+h)} - \red{k\cdot f(x)}} h \end{align*} $$

Step 2

Factor out the $$k$$ and then evaluate the limit.

$$ \begin{align*} g'(x) & = \lim_{x\to h} \frac{\blue k\cdot f(x+h) - \blue k\cdot f(x)} h\\[6pt] & = \lim_{x\to h} \blue k \cdot \frac{f(x+h) - f(x)} h\\[6pt] & = \blue k\cdot \lim_{x\to h} \frac{f(x+h) - f(x)} h\\[6pt] & = \blue k\cdot f'(x) \end{align*} $$

Summary of Rule: $$\displaystyle \frac d {dx}\left(k\cdot f(x)\right) = k\cdot f'(x)$$

Derivatives of Linear Functions

Find $$\displaystyle \frac d {dx}\left(mx + b\right)$$.

Step 1

Evaluate the functions in the definition of the derivative, and simplify.

$$ \begin{align*} \frac d {dx}\left(mx + b\right) & = \lim_{h\to 0} \frac{\blue{f(x+h)}-\red{f(x)}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{m(x+h)+ b}-(\red{mx+b})} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{mx+mh+ b}-\red{mx-b}} h \\[6pt] & = \lim_{h\to 0} \frac{mh} h\\[6pt] & = \lim_{h\to 0} m \end{align*} $$

Step 2

Evaluate the limit.

$$ \frac d {dx}\left(mx + b\right) = \displaystyle\lim_{h\to 0} m = m $$

Summary of Rule: $$\displaystyle \frac d {dx}\left(mx + b\right) = m$$ (The derivative of a linear function is just its slope.)

Derivative of the $$y = \sin kx$$

Find $$\frac d {dx}\left(\sin kx\right)$$.

Step 1

Evaluate the functions in the definition of the derivative.

$$ \begin{align*} \frac d {dx}\left(\sin kx\right) & = \lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\sin(k(x + h))} - \red{\sin kx}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\sin(kx + kh)} - \red{\sin kx}} h \end{align*} $$

Step 2

Expand the sine function using the Sum of Angles for the Sine.

$$ \begin{align*} \frac d {dx}\left(\sin kx\right) & = \lim_{h\to 0} \frac{\sin(\blue{kx} + \red{kh}) - \sin kx} h\\[6pt] & = \lim_{h\to 0} \frac{\sin\blue{kx}\cos\red{kh} + \sin\red{kh}\cos\blue{kx} - \sin kx} h \end{align*} $$

Step 3

Rearrange the numerator so you can factor out the $$\sin kx$$.

$$ \begin{align*} \frac d {dx}\left(\sin kx\right) & = \lim_{h\to 0} \frac{\blue{\sin kx}\cos kh + \sin kh\cos kx - \blue{\sin kx}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\sin kx}\cos kh - \blue{\sin kx} + \sin kh\cos kx} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\sin kx}\left(\cos kh - 1\right) + \sin kh\cos kx} h \end{align*} $$

Step 4

Separate into two limits. Factor out the terms that do not involve $$h$$.

$$ \begin{align*} \frac d {dx}\left(\sin kx\right) & = \lim_{h\to 0} \frac{\blue{\sin kx}\left(\cos kh - 1\right) + \sin kh\,\red{\cos kx}} h\\[6pt] & = \lim_{h\to 0} \left(\frac{\blue{\sin kx}\left(\cos kh - 1\right)} h + \frac{\sin kh\,\red{\cos kx}} h\right)\\[6pt] & = \lim_{h\to 0} \frac{\blue{\sin kx}\left(\cos kh - 1\right)} h + \lim_{h\to 0} \frac{\sin kh\,\red{\cos kx}} h\\[6pt] & = \blue{\sin kx}\cdot \lim_{h\to 0} \frac{\cos kh - 1} h + \red{\cos kx}\cdot \lim_{h\to 0} \frac{\sin kh} h \end{align*} $$

Step 5

Evaluate each limit.

$$ \begin{align*} \frac d {dx}\left(\sin kx\right) & = \sin kx\cdot \blue{\lim_{h\to 0} \frac{\cos kh - 1} h} + \cos kx\cdot \red{\lim_{h\to 0} \frac{\sin kh} h}\\[6pt] & = \sin kx\cdot \blue{(0)} + \cos kx\cdot \red{\lim_{h\to 0} \frac k k \cdot \frac{\sin kh} h}\\[6pt] & = \red k \cos kx\cdot \red{\lim_{h\to 0} \frac{\sin kh} {kh}}\\[6pt] & = k \cos kx\cdot \red{(1)}\\[6pt] & = k \cos kx \end{align*} $$

Summary of Rule: $$\displaystyle \frac d {dx}\left(\sin kx\right) = k\cos kx$$

Derivative of $$y = \cos kx$$

Find $$\frac d {dx}\left(\cos kx\right)$$.

Step 1

Evaluate the functions in the definition of the derivative.

$$ \begin{align*} \frac d {dx}\left(\cos kx\right) & = \lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\cos(k(x + h))} - \red{\cos(kx)}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\cos(kx + kh)} - \red{\cos(kx)}} h \end{align*} $$

Step 2

Expand the cosine function using the Sum of Angles for the Cosine.

$$ \begin{align*} \frac d {dx}\left(\cos kx\right) & = \lim_{h\to 0} \frac{\cos(\blue{kx} + \red{kh}) - \cos(kx)} h\\[6pt] & = \lim_{h\to 0} \frac{\cos\blue{kx}\cos \red{kh} - \sin\blue{kx}\sin\red{kh} - \cos kx} h \end{align*} $$

Step 3

Rearrange the numerator and factor out the $$\cos kx$$.

$$ \begin{align*} \frac d {dx}\left(\cos kx\right) & = \lim_{h\to 0} \frac{\blue{\cos kx}\cos kh - \sin kx\sin kh - \blue{\cos kx}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\cos kx}\cos kh - \blue{\cos kx} - \sin kx\sin kh} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{\cos kx}(\cos kh - 1) - \sin kx\sin kh} h \end{align*} $$

Step 4

Separate into two limits. Factor out the terms not involving $$h$$.

$$ \begin{align*} \frac d {dx}\left(\cos kx\right) & = \lim_{h\to 0} \frac{\blue{\cos kx}(\cos kh - 1) - \red{\sin kx}\sin kh} h\\[6pt] & = \lim_{h\to 0} \left(\frac{\blue{\cos kx}(\cos kh - 1)} h - \frac{\red{\sin kx}\sin kh} h\right)\\[6pt] & = \lim_{h\to 0} \frac{\blue{\cos kx}(\cos kh - 1)} h - \lim_{h\to 0} \frac{\red{\sin kx}\sin kh} h\\[6pt] & = \blue{\cos kx}\cdot\lim_{h\to 0} \frac{\cos kh - 1} h - \red{\sin kx}\cdot\lim_{h\to 0} \frac{\sin kh} h \end{align*} $$

Step 5

Evaluate each limit.

$$ \begin{align*} \frac d {dx}\left(\cos kx\right) & = \cos kx \cdot\blue{\lim_{h\to 0} \frac{\cos kh - 1} h} - \sin kx \cdot\red{\lim_{h\to 0} \frac{\sin kh} h}\\[6pt] & = \cos kx \cdot\blue{(0)} - \sin kx \cdot\red{\lim_{h\to 0} \frac k k\cdot \frac{\sin kh} h}\\[6pt] & = \cos kx \cdot\blue{(0)} - \red k\sin kx \cdot\red{\lim_{h\to 0} \frac{\sin kh} {kh}}\\[6pt] & = - k\sin kx \cdot\red{(1)}\\[6pt] & = - k\sin kx \end{align*} $$

Summary of Rule: $$\displaystyle \frac d {dx}\left(\cos kx\right) = -k\sin kx$$

Derivative of $$y = e^{kx}$$

Find $$\frac d {dx}\left(e^{kx}\right)$$.

Step 1

Evaluate the functions in the definition of the derivative.

$$ \begin{align*} \frac d {dx}\left(e^{kx}\right) & = \lim_{h\to 0} \frac{\blue{f(x + h)} - \red{f(x)}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{e^{k(x+h)}} - \red{e^{kx}}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{e^{kx+kh}} - \red{e^{kx}}} h \end{align*} $$

Step 2

Use the properties of exponents to simplify the numerator.

$$ \begin{align*} \frac d {dx}\left(e^{kx}\right) & = \lim_{h\to 0} \frac{e^{\blue{kx}+\red{kh}} - e^{kx}} h\\[6pt] & = \lim_{h\to 0} \frac{e^{\blue{kx}}\cdot e^{\red{kh}} - e^{kx}} h \end{align*} $$

Step 3

Factor out the $$e^{kx}$$ terms.

$$ \begin{align*} \frac d {dx}\left(e^{kx}\right) & = \lim_{h\to 0} \frac{\blue{e^{kx}}\cdot e^{kh} - \blue{e^{kx}}} h\\[6pt] & = \lim_{h\to 0} \frac{\blue{e^{kx}}\left(e^{kh} - 1\right)} h\\[6pt] & = \blue{e^{kx}}\cdot \lim_{h\to 0} \frac{e^{kh} - 1} h \end{align*} $$

Step 4

Evaluate the limit using the techniques from the lesson on Indeterminate Limits---Exponential Forms.

$$ \begin{align*} \frac d {dx}\left(e^{kx}\right) & = e^{kx}\cdot \lim_{h\to 0} \frac{e^{kh} - 1} h\\[6pt] & = e^{kx}\cdot \lim_{h\to 0}\blue{\frac k k}\cdot \frac{e^{kh} - 1} h\\[6pt] & = \blue k e^{kx}\cdot \lim_{h\to 0}\frac{e^{kh} - 1}{\blue k h}\\[6pt] & = ke^{kx}\cdot \blue{\lim_{h\to 0}\frac{e^{kh} - 1} {kh}}\\[6pt] & = ke^{kx}\blue{(1)}\\[6pt] & = ke^{kx} \end{align*} $$

Summary of Rule: $$\displaystyle \frac d {dx}\left(e^{kx}\right) = ke^{kx}$$

List of Basic Derivatives

Suppose $$f(x)$$ and $$g(x)$$ are differentiable functions and $$k$$ is a constant. Then...

  • Derivative of a Constant: $$\displaystyle \frac d {dx} \left( k \right) = 0$$
  • Derivative of the Identity Function: $$\displaystyle \frac d {dx} \left( x \right) = 1$$
  • Derivative of a Sum: $$\displaystyle \frac d {dx} \left( f(x) + g(x) \right) = f'(x) + g'(x)$$
  • Derivative of a Difference: $$\displaystyle \frac d {dx} \left( f(x) - g(x) \right) = f'(x) - g'(x)$$
  • Derivatives with Constant Coefficients: $$\displaystyle \frac d {dx} \left( k\cdot f(x) \right) = k\cdot f'(x)$$
  • Derivative of a Linear Function: $$\displaystyle \frac d {dx} \left( mx + b \right) = m$$
  • Derivative of the Sine Function: $$\displaystyle \frac d {dx} \left( \sin kx \right) = k\cos kx$$
  • Derivative of the Cosine Function: $$\displaystyle \frac d {dx} \left( \cos kx \right) = -k\sin kx$$
  • Derivative of the Exponential Function: $$\displaystyle \frac d {dx} \left( e^{kx} \right) = ke^{kx}$$

Derivative Rules of Basic Functions. Visual Explanation with color coded examples (2024)

FAQs

How do you visualize the derivative of a function? ›

Derivative Graph Rules

The top graph is the original function, f(x), and the bottom graph is the derivative, f'(x). What do you notice about each pair? If the slope of f(x) is negative, then the graph of f'(x) will be below the x-axis. If the slope of f(x) is positive, then the graph of f'(x) will be above the x-axis.

What are the 7 rules of differentiation with examples? ›

Derivatives Rules Table
Name of the RuleDifferentiation Rule
Constant Multiple Ruled/dx (c f(x)) = c d/dx (f(x))
Sum/Difference Ruled/dx (f(x) ± g(x)) = d/dx (f(x)) ± d/dx (g(x))
Product Ruled/dx (f(x)) · g(x)) = f(x) d/dx (g(x)) + g(x) d/dx (f(x))
3 more rows

What are the basic derivative rules? ›

Derivative rules
Derivative sum rule( a f (x) + bg(x) ) ' = a f ' (x) + bg' (x)
Derivative product rule( f (x) ∙ g(x) ) ' = f ' (x) g(x) + f (x) g' (x)
Derivative quotient rule
Derivative chain rulef ( g(x) ) ' = f ' ( g(x) ) ∙ g' (x)

How to take a derivative example? ›

To find the derivative of h(x), take the derivative of each term. The derivative of x 2 is 2x, and the derivative of 6 is 0 so h(x) = 2x. In this example, unlike in Example 2, the constant is not attached to the variable that is being differentiated, so its derivative is 0.

How is a derivative shown on a graph? ›

The derivative graph rules are as follows: If the function is increasing with positive slopes, the derivative is positive and plotted above the x-axis. If the function is decreasing with negative slopes, the derivative is negative and plotted below the x-axis.

What are the 4 types of differentiation? ›

You can differentiate instruction across four main areas: content, process, product, and environment.

What are the 4 principles of differentiation? ›

Dr. Carol Tomlinson proposed that there are four ways to differentiate instruction: through content, process, product, and learning environment (2003; Tomlinson & Imbeau, 2010).

What is the derivative for dummies? ›

The derivative of a function describes the function's instantaneous rate of change at a certain point. Another common interpretation is that the derivative gives us the slope of the line tangent to the function's graph at that point.

What are the basic concepts of derivatives in math? ›

Derivatives are defined as the varying rate of change of a function with respect to an independent variable. The derivative is primarily used when there is some varying quantity, and the rate of change is not constant.

What is an example of a derivation rule? ›

A derivation rule is a rule by which values of a particular type can be derived or calculated: for example, the cost of an order line is the product unit cost multiplied by the order quantity, in turn multiplied by 100% minus the discount rate (if any).

What are the 5 examples of derivatives? ›

Five of the more popular derivatives are options, single stock futures, warrants, a contract for difference, and index return swaps. Options let investors hedge risk or speculate by taking on more risk. A stock warrant means the holder has the right to buy the stock at a certain price at an agreed-upon date.

What is derivative and examples? ›

Derivatives are securities whose value is dependent on or derived from an underlying asset. For example, an oil futures contract is a type of derivative whose value is based on the market price of oil.

What is a common example of derivative work? ›

A derivative work is a work based on or derived from one or more already exist- ing works. Common derivative works include translations, musical arrange- ments, motion picture versions of literary material or plays, art reproductions, abridgments, and condensations of preexisting works.

How is the derivative of a function represented? ›

The derivative of a function is defined as the instantaneous rate of change of a function at a specific point. The derivative gives the exact slope along the curve at a specific point. The derivative of the function is represented as d y d x , which means the derivative of with respect to the variable .

How to find the graph of a function from its derivative? ›

If the function, f(x) has a positive slope, then the derivate f'(x) is graphed above the x-axis. If the function, f(x) has a negative slope, then the derivate f'(x) is graphed below the x-axis. All extremas will become x-intercepts, within the derivative graph.

How to draw the derivative graph from the original? ›

The thing to remember is the derivative is the slope of the line tangent to the graph. So pick a few points on the graph, and sketch the tangent line. If the slope is positive, it corresponds to a point on the graph of the derivative that's above the x-axis; if the slope is negative, the point is below the x-axis.

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