Answer :
We have to find the derivative of y = x^(-2) using the first principles. This means that we have to calculate the derivative from the limit of the secant line.
We can then write:
[tex]\begin{gathered} f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\ f^{\prime}(x)=\lim_{h\to0}\frac{(x+h)^{-2}-x^{-2}}{h} \end{gathered}[/tex]We can rearrange the expression first and then calculate the limit:
[tex]\begin{gathered} \frac{1}{h}[\frac{1}{(x+h)^2}-\frac{1}{x^2}] \\ \frac{1}{h}[\frac{x^2-(x+h)^2}{(x+h)^2x^2}] \\ \frac{1}{h}(\frac{x^2-x^2-2xh-h^2}{(x+h)^2x^2}) \\ \frac{1}{h}(\frac{-2xh-h^2}{(x+h)^2x^2}) \\ \frac{-2x-h}{(x+h)^2x^2} \end{gathered}[/tex]We can now calculate the limit as:
[tex]\lim_{h\to0}\frac{-2x-h}{(x+h)^2x^2}=\frac{-2x-0}{(x+0)^2x^2}=\frac{-2x}{x^4}=-\frac{2}{x^3}[/tex]Answer: the first derivative is -2/x³