Find the derivative of $h(x) = 1/x^2$ by definition

Question

Can someone please explain the process of finding the derivative $h'(x)$ of $h(x) = \dfrac{1}{x^2} $

using the delta ($\Delta$) notation.

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What I managed:

1. $h '(x) \displaystyle = \lim _{\Delta x\rightarrow 0} \frac{h(x+\Delta x)- h(x)}{\Delta x}$
   
   
2. $h '(x) \displaystyle = \lim _{\Delta x\rightarrow 0} \frac{1/(x+\Delta x)^2 - 1/x^2}{\Delta x}$
   
   
3. $h '(x) \displaystyle = \lim _{\Delta x\rightarrow 0} \frac{x^2-(x+\Delta x)^2}{(x+\Delta x)^2x^2\Delta x}$

And after this point, I do not know how to proceed.

The next steps have been given as:

4. $h '(x) \displaystyle = \lim _{\Delta x\rightarrow 0} \frac{-2x\Delta x - (\Delta x)^2}{(x+\Delta x)^2x^2\Delta x}$
   
   

5. $h '(x) \displaystyle = \lim _{\Delta x\rightarrow 0} \frac{-2x - \Delta x}{(x+\Delta x)^2x^2}$
   
   

With the answer being $h'(x) = -\dfrac{2}{x^{3}}$

Can someone please explain the process between steps 3 and 4?

Answer 1

Note that $\displaystyle x^2-(x+\Delta x)^2=x^2-(x^2+2x\Delta x+(\Delta x)^2)=-2x\Delta x-(\Delta x)^2$.

Answer 2

just simplify the equation like : $(a + b)^2 = a^2 + 2ab + b^2 => x^2 - (x +Δx) = x^2 - (x^2 + 2xΔx + Δx^2) = -2xΔx - Δx^2$

Answer 3

$$\frac{h(x+\Delta{x})-h(x)}{\Delta{x}}=\frac{\frac{1}{(x+\Delta{x})^2}-\frac{1}{x^2}}{\Delta{x}}=\frac{\frac{1}{x^2+2x\Delta{x}+\Delta{x^2}}-\frac{1}{x^2}}{\Delta{x}}=\frac{-2x\Delta{x}-\Delta{x^2}}{(\Delta{x})(x^2)(x^2+2x\Delta{x}+\Delta{x^2})}=\frac{-2x-\Delta{x}}{x^4+2x^3\Delta{x}+x^2\Delta{x^2}}$$ and since $\Delta{x}$ tends to 0, this equals $$\frac{-2x}{x^4}=\frac{-2}{x^3}$$