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Find the derivative of $$ f(x) = 3x^{-\frac{1}{3}} $$ using first principles and difference of cubes.

My teacher has hinted that $$ \frac{3(x+h)^{-\frac{1}{3}} - 3x^{-\frac{1}{3}}}{h} $$ can be turned into $$ \frac{{(x+h)^{-\frac{1}{3}} + x^{-\frac{1}{3}}}}{h} $$ with some factoring, (the entire fraction must be multiplied by 3?), but I can't see how he got there unless there is a typo with "+" actually being "-". Also he hinted that from the equation above we can multiply by the conjugate of the numerator and use the difference of cubes to solve the limit but I've been playing around with that for the past couple hours and still can't see how to get to $$ -\frac{1}{x^{4/3}} $$

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Let $f(x)=x^{-\frac{1}{3}}$

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(x+h)^{-1/3} - x^{-1/3}}{h}= \lim_{h \to 0} \frac{x^{1/3}-(x+h)^{1/3}}{h\times x^{1/3}\times (x+h)^{1/3}}=$$ $$=\lim_{h \to 0} \frac{x^{1/3}-(x+h)^{1/3}}{h\times x^{1/3}\times (x+h)^{1/3}} \times \frac{x^{2/3} + x^{1/3}\times (x+h)^{1/3} +(x+h)^{2/3}}{x^{2/3} + x^{1/3}\times (x+h)^{1/3} +(x+h)^{2/3}}=$$ $$=\lim_{h \to 0} \frac{x-(x+h)}{h\times x^{1/3}\times (x+h)^{1/3} \times (x^{2/3} + x^{1/3}\times (x+h)^{1/3} +(x+h)^{2/3})}=$$ $$=\lim_{h \to 0} \frac{-1}{x^{1/3}\times (x+h)^{1/3} \times (x^{2/3} + x^{1/3}\times (x+h)^{1/3} +(x+h)^{2/3})} =$$ $$= \frac{-1}{x^{1/3}\times x^{1/3} \times (x^{2/3} + x^{1/3}\times x^{1/3} +x^{2/3})}=\frac{-1}{3x^{4/3}}$$

If $f(x)=3x^{-\frac{1}{3}}$ then $$f'(x)=3\times \frac{-1}{3x^{4/3}}=\frac{-1}{x^{4/3}}$$

Lion Heart
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The difference of cubes formula says $$a^3 - b^3 = (a-b)(a^2 + ab + b^2).$$ I will just choose to find the derivative of $f(x) = x^{-1/3}$ so we are dealing with the limit $$f'(x) := \lim_{h \to 0} \frac{(x+h)^{-1/3} - x^{-1/3}}{h}.$$ Your teacher is asking you to make a judicious choice of $a,b$ and match terms so that we can simplify our problem. Note that we don't know how to deal with the differences of cube roots, but if we cubed the quantities in the numerator, we'd have an easier time. Hopefully this helps you figure out what $a$ and $b$ should be.

1mdlrjcmed
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