Question

given that y = 2x^-2, find dy/dx from first principles

Answer 1

bring the exponent to the front and multiply by the 2, then subtract 1 from the exponent.

dy/dx = -4x^-3

Answer 2

no, from first princicples means the whole
dy/dx=lim(h->0) [f(x+h)-f(x)]/h

Answer 3

you did the short way

Answer 4

Oh, woops. I forgot about that.

Answer 5

I haven't used that since calc 1

Answer 6

\[\begin{align*}
\frac{dy}{dx}&=\lim \limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\
&=\lim \limits_{\Delta x\to 0}\frac{2(x+\Delta x)^{-2}-2x^{-2}}{\Delta x}\\
&=2\lim \limits_{\Delta x\to 0}\frac{x^{2}-(x+\Delta x)^2}{x^2(\Delta x )(x+\Delta x)^2}\\
&=\frac{2}{x^2}\lim \limits_{\Delta x\to 0}\frac{-\Delta x(2x+\Delta x)}{(\Delta x )(x+\Delta x)^2}\\
&=\frac{2}{x^2}\lim \limits_{\Delta x\to 0}\frac{-(2x+\Delta x)}{(x+\Delta x)^2}\\
&=\frac{2}{x^2}\frac{-2x}{x^2}\\
&=-4x^{-3}
\end{align*}\]