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# 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 $$\displaystyle y = \frac{1}{x}\,$$, $$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x^2} = -x^{-2}$$