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 \(a\).
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.
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}