Rules of Differentiation for Finding Derivatives

The good news is that you don’t have to differentiate all expressions with the definition of the derivative. Here are all the rules you can follow to be able to differentiate the basic functions.

Look closely at the examples, and make sure you understand what is happening.

Rule

Rules for Derivation

\begin{array}{l} f (x) & = k & \Leftrightarrow & f^{'} (x) & = 0 \\ f (x) & = a x & \Rightarrow & f^{'} (x) & = a \\ f (x) & = x^{n} & \Rightarrow & f^{'} (x) & = n x^{n - 1} \\ f (x) & = \sqrt{x} = x^{\frac{1}{2}} & \Rightarrow & f^{'} (x) & = \frac{1}{2} x^{- \frac{1}{2}} \\ = \frac{1}{2 \sqrt{x}} \\ f (x) & = e^{x} & \Rightarrow & f^{'} (x) & = e^{x} \\ f (x) & = e^{k x} & \Rightarrow & f^{'} (x) & = k e^{k x} \\ f (x) & = a^{x} & \Rightarrow & f^{'} (x) & = a^{x} \ln a \\ f (x) & = \ln x & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \ln (k x) & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \sin x & \Rightarrow & f^{'} (x) & = \cos x \\ f (x) & = \cos x & \Rightarrow & f^{'} (x) & = - \sin x \\ f (x) & = \sin (k x) & \Rightarrow & f^{'} (x) & = k \cos (k x) \\ f (x) & = \cos (k x) & \Rightarrow & f^{'} (x) & = - k \sin (k x) \\ f (x) & = \tan x & \Rightarrow & f^{'} (x) & = \frac{1}{\cos^{2} x} \\ = \tan^{2} x \\ f (x) & = \tan (k x) & \Rightarrow & f^{'} (x) & = \frac{k}{\cos^{2} (k x)} \\ = k + k \tan^{2} (k x) \end{array}

\begin{array}{l} f (x) & = k & \Leftrightarrow & f^{'} (x) & = 0 \\ f (x) & = a x & \Rightarrow & f^{'} (x) & = a \\ f (x) & = x^{n} & \Rightarrow & f^{'} (x) & = n x^{n - 1} \\ f (x) & = \sqrt{x} = x^{\frac{1}{2}} & \Rightarrow & f^{'} (x) & = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}} \\ f (x) & = e^{x} & \Rightarrow & f^{'} (x) & = e^{x} \\ f (x) & = e^{k x} & \Rightarrow & f^{'} (x) & = k e^{k x} \\ f (x) & = a^{x} & \Rightarrow & f^{'} (x) & = a^{x} \ln a \\ f (x) & = \ln x & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \ln (k x) & \Rightarrow & f^{'} (x) & = \frac{1}{x} \\ f (x) & = \sin x & \Rightarrow & f^{'} (x) & = \cos x \\ f (x) & = \cos x & \Rightarrow & f^{'} (x) & = - \sin x \\ f (x) & = \sin (k x) & \Rightarrow & f^{'} (x) & = k \cos (k x) \\ f (x) & = \cos (k x) & \Rightarrow & f^{'} (x) & = - k \sin (k x) \\ f (x) & = \tan x & \Rightarrow & f^{'} (x) & = \frac{1}{\cos^{2} x} = \tan^{2} x \\ f (x) & = \tan (k x) & \Rightarrow & f^{'} (x) & = \frac{k}{\cos^{2} (k x)} = k + k \tan^{2} (k x) \end{array}

For functions

f (x)

g (x)

and constants

k

, the following applies:

\begin{array}{l} {(f (x) \pm g (x))}^{'} & = f^{'} (x) \pm g^{'} (x) \\ {(k f (x))}^{'} & = k f^{'} (x) \end{array}

In addition to these rules, you can read the entries about the chain rule, the product rule and the quotient rule.

Below are some examples to help you practice these rules.

Example 1

f (x) = 6 \Rightarrow f^{'} (x) = 0

Example 2

f (x) = 12 x \Rightarrow f^{'} (x) = 12

Example 3

f (x) = x^{3} \Rightarrow f^{'} (x) = 3 x^{2}

Example 4

f (x) = 7 x^{5} \Rightarrow f^{'} (x) = 35 x^{4}

Example 5

f (x) = 2 x^{3} - 6 x^{9} \Rightarrow f^{'} (x) = 6 x^{2} - 54 x^{8}

f (x) = 2 x^{3} - 6 x^{9} \Rightarrow f^{'} (x) = 6 x^{2} - 54 x^{8}

Example 6

f (x) = \frac{1}{x} = x^{- 1} \Rightarrow f^{'} (x) = - \frac{1}{x^{2}}

Example 7

f (x) = e^{4 x} \Rightarrow f^{'} (x) = 4 e^{4 x}

Example 8

f (x) = 3^{x} \Rightarrow f^{'} (x) = 3^{x} \ln 3

Example 9

f (x) = \ln 4 x \Rightarrow f^{'} (x) = \frac{1}{x}

Example 10

Jax Teller is driving along a straight country road in Italy. The number $s$ of kilometers Jax has covered after $t$ hours is given by the function $s (t) = 50 t^{2}$ . What can you say about his speed?

You find Jax’s speed after $t$ hours by differentiating the function $s (t)$ :

s^{'} (t) = 50 \cdot 2 t = 100 t

According to the model, Jax increases his speed by $100$ km/h every hour.

After 30 minutes ( $0.5$ hours), his speed is

s^{'} (0.5) = (100 \cdot 0.5) km/h = 50 km/h

After 1 hour, his speed is

s^{'} (1) = (100 \cdot 1) km/h = 100 km/h

After $2.5$ hours, his speed is

s^{'} (2.5) = (100 \cdot 2.5) km/h = 250 km/h

When you drive a long time in a straight direction, you can start experiencing speed blindness, which means you don’t notice that you’re driving too fast!

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How Do You Determine if a Function Is Differentiable?

How Does the Chain Rule for Differentiation Work?

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