### Differentiation: The derivative

### The notion of derivative

The derivative is a function that indicates for each point #x# what the slope is at that point. In other words, a function that assigns the slope of the tangent line to a point #x#.

The slope

The **slope** of a function #\blue{f}# at a point #a# can be found by calculating the difference quotient for #[a,a+\orange{h}]# and letting #\orange{h}# approach zero. We write this as follows:

\[\orange{h}\to0\]

If we do not determine the difference quotient at a point but for a variable # x #, we get the **derivative** #\blue{f}#. We denote the derivative with #\green{f'}#.

For the example on the right, it is only indicated in the second to last step that #\orange{h} \to 0#. However, this should be at every step, but we have omitted it for the sake of convenience.

**Example**

\[\begin{array}{rcl}

\blue{f(x)}&=&\blue{x^2} \\

\green{f'(x)}&=&\dfrac{\blue{(}x+\orange{h}\blue{)^2}-\blue{x^2}}{\orange{h}}\\&=&\dfrac{\blue{x^2}+2x\cdot \orange{h}+\orange{h}^2-\blue{x^2}}{\orange{h}}\\&=&\dfrac{2x\cdot \orange{h}+\orange{h}^2}{\orange{h}}\\&=&2x+\orange{h} \quad \text{with} \quad \orange{h} \to 0\\&=& \green{2x} \end{array}\]

We call #f'# the **derivative** of #f#.

The derivative

The **derivative** of a function #\blue{f}# is denoted as #f'#:

\[f'(x)=\dfrac{\blue{f(}x+\orange{h}\blue{)}-\blue{f(}x\blue{)}}{\orange{h}} \quad \text{with} \quad \orange{h}\to 0\]

Calculating the derivative of a function #f# is called **diferentiation** of #f#.

Not every function can be differentiated. A function of which we can determine the derivative is called a **differentiable** function. In this course will will only deal with functions that are differentiable.

When we write #\frac{\dd}{\dd x}f# or #\frac{\dd f}{\dd x}#, we mean #f'#; these three all mean the same thing.

For #h \to 0#, we find:

\[\begin{array}{rcl} f'(x)&=&\dfrac{f(x+h)-f(x)}{h}\\ && \blue{\text{definition derivative}}\\ &=& \dfrac{5(x+h)^3 + 5(x+h) - (5x^3+5x)}{h}\\ &&\blue{\text{substituted}}\\ &=& \dfrac{5x^3 + 15x^2\cdot h + 15x\cdot h^2 + 5h^3 + 5x + 5h -5x^3-5x}{h} \\ && \blue{\text{expanded brackets}}\\&=& \dfrac{15x^2\cdot h + 15x\cdot h^2 + 5h^3 +5h }{h} \\&&\blue{5x^3 \text{ and } -5x^3, \text{ and }5x \text{ and } -5x \text{ cancel each other out}}\\ &=& 15x^2 + 15x\cdot h + 5h^2 +5 \\ &&\blue{\text{eliminate }h} \\&=& 15x^2 + 5

\\ && \blue{\text{let }h \text{ approach }0}\end{array}\]

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