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1. Definition of Derivative

Definition

Definition 1: The derivative of a function $$f$$ at a point $$a$$ is defined as $\lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}$

Definition 2: The derivative of a function $$f$$ at a point $$a$$ is defined as $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$

The derivative of a function $$f(x)$$ is the function $$f'(x)$$ defined by $$f'(a)$$ is equal to the derivative of $$f$$ at the point $$a$$ for every point $$a$$.

Explanation

The derivative of $$f(x)$$ at the point $$a$$ is the slope of the tangent line of $$f$$ at $$a.$$

Consider the graph $$f(x) = \frac{x^3}{10}+1.$$

The slope between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is $\frac{y_2 - y_1}{x_2 - x_1}$ Given a coordinate $$x,$$ the $$y$$ value in the graph of $$f$$ is $$f(x).$$ So, the slope between two points on the graph is $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ Fix the $$x$$-value $$x = 1.$$ The point on the graph is $$(1, f(1)).$$

The slope between the points $$(1,f(1))$$ and $$(2,f(2))$$ can be computed using the slope formula: \begin{align} \frac{f(2)-f(1)}{2-1} & = f(2)-f(1) \\ & = \frac{8}{10}+1-(\frac{1}{10}-1) \\ & = \frac{7}{10} \end{align}

In the above paragraph, the difference is the derivative formula without then limit when $$h = 1.$$ The slope is $\frac{f(1+1)-f(1)}{(1+1)-1}$

When $$h = \frac{1}{2},$$ we get $$x + h = 1 + \frac{1}{2} = \frac{3}{2}.$$ The slope between $$(1, f(1))$$ and $$\left(\frac{3}{2}, f\left(\frac{3}{2}\right)\right)$$ is \begin{align} \frac{f\left(\frac{3}{2}\right)-f(1)}{\frac{3}{2}-1} & = 2\left(f\left(\frac{3}{2}\right)-f(1)\right) \\ & = 2\left(\frac{27}{80}+1-\frac{1}{10}-1\right) \\ & = \frac{19}{40} \end{align}

We can find the slope between the points $$(1, f(1))$$ and $$(1+h, f(1+h))$$ for smaller and smaller $$h.$$ As $$h$$ approaches $$0,$$ the slope will approach the slope of the tangent line, which is \begin{align} \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} & = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{(1+h)^3}{10}+1-\frac{1}{10}-1\right) \\ & = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{1+3h+3h^2+h^3}{10} - \frac{1}{10} \right) \\ & = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{3h+3h^2+h^3}{10} \right) \\ & = \lim_{h \rightarrow 0} \frac{3+3h+h^2}{10} \\ & = \frac{3}{10} \\ \end{align}

If you look at the lines point by point, you can see how the lines approach the tangent in the limit.

Example

The derivative of $$f(x) = 4$$ is $$f'(x) = 0$$ at every point $$x.$$

For any point $$x,$$ \begin{align} \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} & = \lim_{h \rightarrow 0}\frac{4-4}{h} \\ & = \lim_{h \rightarrow 0} 0 \\ & = 0 \end{align}

Claim: If $$f(x) = c$$ for any constant $$c,$$ then $$f'(x) = 0$$ at every point $$x.$$ This can be shown by replacing $$4 - 4$$ with $$c - c.$$

The derivative of $$f(x) = x$$ is $$f'(x) = 1$$ at every point $$x.$$

For any point $$x,$$ \begin{align} \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} & = \lim_{h \rightarrow 0}\frac{x+h-x}{h} \\ & = \lim_{h \rightarrow 0} 1 \\ & = 1 \end{align}

The derivative of $$f(x) = x^2$$ is $$f'(x) = 2x$$ for every point $$x.$$

For any point $$x,$$ \begin{align} \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} & = \lim_{h \rightarrow 0}\frac{(x+h)^2-x^2}{h} \\ & = \lim_{h \rightarrow 0} \frac{x^2+2xh+h^2 - x^2}{h} \\ & = \lim_{h \rightarrow 0} \frac{2xh+h^2}{h} \\ & = \lim_{h \rightarrow 0} 2x + h \\ & = 2x \end{align}