Finding derivative $f(x)={2\over x^3}$ I have to find the derivative and the slope at $a=6$
The function is  $f(x)={2\over x^3}$
I have to find the answer using the formula,
$$f'(x)= \lim_{\Delta x \to 0} {f(x+ \Delta x) - f(x) \over \Delta x}$$
I tried getting rid of the denominator, but I think I'm getting mixed up somewhere.
The answer book says the slope is ${-1 \over 216}$
Here's what I've done,
$$f'(x)= \lim _{\Delta x\to 0} {f(x +\Delta x) - f(x)\over \Delta x} = \lim_{\Delta x \to 0} {{2 \over (x+\Delta x)^3} - {2 \over x^3}\over\Delta x}
\\ = \lim_{\Delta x \to 0} {({2 \over (x+\Delta x)^3} - {2 \over x^3}) ((x+\Delta x)^3(x^3))\over\Delta x ((x+\Delta x)^3(x^3))}
\\\\= \lim_{\Delta x \to 0} {2(x^3) - 2(x+\Delta x)^3\over \Delta x (x+\Delta x)^3(x^3)}$$
To cancel out the denominator.
Did I do this right??
Should I now expand the parentheses and cancel things out??
Would I still get the same answer as the answer book??(Because the answer book have made a typo once before).
Thanks

You're absolutely right! 
$$\begin{align}f'(x) &= \lim_{\Delta \to 0} \frac{2x^3 - 2(x^3 + 3x^2\Delta x + 3x \Delta^2 x + \Delta^3 x )}{\Delta x (x + \Delta x)^3(x^3)} \\&= \lim_{\Delta x \to 0} \frac{-6x^2 - 6x\Delta x - 2\Delta ^3 x }{(x + \Delta x)^3(x^3)} \\&= \lim_{\Delta x \to 0} \frac{-x^2\Big(6 + 6\frac{\Delta x}{x} + 2\frac{\Delta ^3 x}{x^2}\Big) }{(x + \Delta x)^3(x^3)} \\&=\lim_{\Delta x \to 0} \frac{-\Big(6 + 6\frac{\Delta x}{x} + 2\frac{\Delta ^3 x}{x^2}\Big) }{(x + \Delta x)^3x} \\&=  -\frac{6}{x^4}\end{align}$$
