Finding derivative $f(x)={2\over x^3}$

Question

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

Answer 1

You're absolutely right!

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

Answer 2

Yes, so far this is correct. Now you should expand the parenthesis and cancel the $2x^3$ first. Then you can cancel the $\Delta x$. You should then be able to get the answer from the book.