Differentiation first principles for cube Find the derivative of function $f(x) = \sqrt{x} + \dfrac{1}{x^3}$ from the first principles.
I tried to use the formula
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
$$
I try to expand for the $x^3$ first but it looks more and more complicated. The answer end up with $0$.
 A: Hint: (I'm assuming you mean $\require{cancel}f(x) = \sqrt{x} + \frac1{x^3}$.) Here's the "hard" computation for a few similar functions. You can see the pattern: in the end, there is a factor of $h$ in the numerator which will cancel with the $h$ that you are going to use as the denominator in the difference quotient, and all but the first term of the remaining factor will vanish as $h\to 0$.
For $f(x) = \frac1x$:
$$\tfrac1{(x+h)}-\tfrac1{x} = \tfrac x{x(x+h)}-\tfrac{x+h}{x(x+h)}=\tfrac{\cancel{x}-(\cancel{x}+h)}{x(x+h)} = \tfrac{-h}{x(x+h)}$$
For $f(x)=\frac1{x^2}$:
$$\tfrac1{(x+h)^2}-\tfrac1{x^2} = \tfrac {x^2}{x^2(x+h)^2}-\tfrac{(x+h)^2}{x^2(x+h)^2}=\tfrac{\cancel{x^2}-(\cancel{x^2}+2hx + h^2)}{x^2(x+h)^2}=\tfrac{-2xh - h^2}{x^2(x+h)^2}=\tfrac{-h(2x + h)}{x^2(x+h)^2}$$
For $f(x) = \frac1{x^3}$ (which is what you are using):
$$\tfrac1{(x+h)^3}-\tfrac1{x^3} = \tfrac {x^3}{x^3(x+h)^3}-\tfrac{(x+h)^3}{x^3(x+h)^3}=\tfrac{\cancel{x^3}-(\cancel{x^3}+3hx^2 + 3h^2x + h^3)}{x^3(x+h)^3}=\tfrac{-3hx^2 - 3h^2x - h^3}{x^3(x+h)^3}=\tfrac{-h(3x^2 + 2hx + h^2)}{x^3(x+h)^3}$$
I presume you can handle the $\sqrt{x}$ term without difficulty.
A: And for $\sqrt$, there is the standard conjugating approach:
$\sqrt{x+h}-\sqrt{x}
=(\sqrt{x+h}-\sqrt{x})\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}
=\frac{(x+h)-x}{\sqrt{x+h}+\sqrt{x}}
=\frac{h}{\sqrt{x+h}+\sqrt{x}}
\to \frac{h}{\sqrt{x}}
$
