Differentiation Rules

Niamh Ryan
Note by Niamh Ryan, updated more than 1 year ago
Niamh Ryan
Created by Niamh Ryan almost 4 years ago


The next resource in our series on differentiation for high school students looks at differentiation rules. These notes contain explanations on differentiating x\n and differentiating functions. Examples of each type are presented in the form of solved equations.

Resource summary

Page 1

Differentiating \(x^n\)

This is the one of the most frequently used rules of differentiation.   \[ \text{If }y=x^n \text{ then} \frac{dy}{dx} = nx^{n-1}\]   This can be used to differentiate surds or fractions involving powers of \( x\) .   For example, \( \sqrt{x} \) can also be written as \( x^{\frac{1}{2}}\) , allowing it to be easily differentiated using the above rule.   \( \frac{3}{x}\) can also be written as \( 3x^{-1}\) , so    \[\frac{d}{dx}(\frac{3}{x})=-3x^{-2}\]  

Page 2

Differentiating functions

\(\frac{d}{dx} \) is a linear operator.  This means it can be separated out over addition or subtraction and constants can be moved outside the operation. Therefore,  \[\frac{d}{dx}(x^m + x^n) = \frac{d}{dx}(x^m) +\frac{d}{dx}(x^n)\]   and \[ \frac{d}{dx}(kx^m) = k\frac{d}{dx}(x^m)\] where \(k\) is any constant.

Page 3

Example 1

Example 1: Differentiate the following function: \[y=3\sqrt{x} + \frac{4}{x^2}\] Answer: As demonstrated on the previous page, we can treat the two terms in this function as if they were separate functions and their constants do not change with differentiation. First, rewrite the function using exponents instead of surds and fractions: \[ y=3x^{\frac{1}{2}} + 4x^{-2}\] and then differentiatie:\[\frac{dy}{dx} = 3(\frac{1}{2})x^{\frac{1}{2} - 1}+4(-2)x^{-2-1}= \frac{3}{2}x^{-\frac{1}{2}}-8x^{-3}\] This can be simplified to give: \[\frac{dy}{dx}=\frac{3}{2 \sqrt{x}} - \frac{8}{x^{-3}}\]

Page 4

Example 2

Example 2: Differentiate the following function: \[y=\frac{x^{2}-3x-4}{x+1}\] Answer: Where there's a fraction with \(x\) terms on top and bottom, factorise the top before continuing. This gives:\[y=\frac{(x-4)(x+1)}{x+1} \] Clearly \((x+1)\) can be factored from top and bottom to give\[y=x-4\] which is the same as \[y=x^1-4x^0\] and then differentiate:\[\frac{dy}{dx} = 1\]  

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