Does the derivative of $x^{-1}$ and of $x^3-x$ equal $-\frac{1}{x^{2}}$ and $3x^2-1$?

Question

I want to check my answers concerning the derivative of the following functions: $\displaystyle f(x)= \frac{1}{x}$ and of $\displaystyle j(x)= x^3-x$

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$$\displaystyle f(x)= \frac{1}{x}$$ $$\begin{align} f'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ & = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \\ & = \lim_{h \to 0} \frac{\frac{x}{x(x+h)} - \frac{(x+h)}{x(x+h)}}{h} \\ & = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x^2+hx}}{h} \\ & = \lim_{h \to 0} \frac{\frac{-h}{x^2+hx}}{h} \\ & = \lim_{h \to 0} \frac{-h}{x^2+hx} \times \frac{1}{h} \\ & = \lim_{h \to 0} \frac{-1}{x^2+hx} \\ & = -\frac{1}{x^2} \end{align}$$

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$$j(x) = x^3 - x$$ $$\begin{align} j'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ & = \lim_{h \to 0} \frac{(x+h)^3 - x^3 - x}{h} \\ & = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x - h - x^3 + x}{h} \\ & = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 - h}{h} \\ & = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 - 1)}{h} \\ & = \frac{3x^2 - 1}{1} \\ & = 3x^2 - 1 \end{align}$$

Please, be rude, if you see any error correct me.

Answer 1

The title of your post says the derivative of $\frac 1x$ is $\frac{1}{x^2}$ but in your post, you found that the derivative is $-\frac{1}{x^2}$, which is correct. Also, in the title you wrote that the derivative of $x^3-x$ is $3x^2$, but you found that it is $3x^2-1$

Your calculations are correct, keep up the good work!

Answer 2

Your title is wrong but your limits in the actual post are correct.