# Find $\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\sqrt[3]{x+2}}$ using only the definition of the derivative

I am trying to find $$\frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{\sqrt[3]{x+2}}$$ using only the definition of the derivative.

I have gotten to this point $$\lim_{h \to 0}\;\dfrac{\dfrac{1}{\sqrt[3]{x+2}}+\dfrac{1}{\sqrt[3]{x+h+2}}}{h(\sqrt[3]{x^2+xh+4x+2h+4})}$$ by finding a common denominator and then expanding. I'm not sure where to go from here, or if I should try using $\lim_{x \to a}\frac{f(x)-f(a)}{x-a}$ instead.

I have looked at Finding derivative of $\frac{1}{\sqrt{x+2}}$ using only the definition of the derivative $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ and the link within that page, and am also wondering if the solution is similar to that one, being that the only difference is a cube root instead of a square root.

• You can do this as letting $\frac{1}{x+2}=y^3$ and then differentiating both sides with respect to x using definition of derivatives to arrive at the value of $y'$ (if it is allowed) and I don't think this disturb the condition of question – Atul Mishra Oct 19 '17 at 5:27

Using the limit definition of derivative: $$f'(x) = \lim_{h\to 0} \frac{(x+2+h)^{-1/3} - (x+2)^{-1/3}}{h}$$
Now let $x+2+h = u^3$ and $x+2 = v^3$. So now we have as $h \to 0$, $u^{3}-v^3 \to 0$. So $u\to v$
\begin{align} \lim_{u\to v} \frac{u^{-1}-v^{-1}}{u^{3}-v^{3}} &= \lim_{u\to v}\frac{-1}{uv(u^2+uv+v^2)}\\ &= \frac{-1}{v^2(v^2+v^2+v^2)}\\ &= \frac{-1}{3v^4} \\ &=\frac{-1}{3 } (x+2)^{-4/3} \end{align}