What is calculus?
Example using first principle to find the derivative
Use algebraic methods to find a formula for the derivative of the function \(\displaystyle y = \frac{1}{x}\)
The slope between two points at \(x\) and \(x + h\) is
\begin{eqnarray*}
\frac{\Delta f}{\Delta x} &=&
\frac{f(x + h) - f(x)}{h}\\
&=& \displaystyle\frac{\displaystyle\frac{1}{x + h} - \displaystyle\frac{1}{x}}{h}\\
&=& \frac{1}{h}\left(\frac{1}{x + h} -
\frac{1}{x}\right)\\
&=& \frac{1}{h}\left(\frac{x - (x + h)}{x(x +
h)}\right)\\
&=& \frac{1}{h}\left(\frac{-h}{x^2 + xh}\right)\\
&=& \frac{-1}{x^2 + xh}
\end{eqnarray*}
Thus, the limiting value as \(h \rightarrow 0\) is \(\displaystyle{-\frac{1}{x^2}}\)
That is, for \(\displaystyle y = \frac{1}{x}\,\), \(\displaystyle
\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x^2} = -x^{-2}\)
To do
- Derivative as a limit activity (Khan Academy)
More info
- Differentiation from first principles teach yourself guide (mathcentre)
- The derivative of \(f(x)=x^2\) for any \(x\) video (Khan Academy)