I had to solve an exercise and was going to ask for help here, but as I typed, I spotted an error myself. I felt sorry to throw away all that MathJax stuff I typed for more than an hour, so I decided to post the question anyway, since it still can be useful to someone. If someone has any suggestions, corrections, I would be very grateful.
This is the task itself, which I'd like to quote, at least for search engines to index, if someone else has this problem.
Graph the curves in Exercises 39-48.
a. Where do the graphs appear to have vertical tangent lines?
b. Confirm your findings in part (a) with limit calculations …
The problem itself:
I can't solve exercise 45: $y = x^{2/3} - (x - 1)^{1/3}$.
I would be grateful if someone told me how to solve this exactly, but any help or hint will be greatly appreciated.
I need to find the derivative of this function, but by limit calculations. I know, the derivative should be $\frac{2}{3x^{1/3}} - \frac{1}{3(x- 1)^{2/3}}$, but I can't find it through the limit:
$$ \lim_{h \to 0} \frac{(x + h)^{2/3} - (x - 1 + h)^{1/3} - x^{2/3} + (x- 1)^{1/3}}{h} $$
Tried approach 1:
I tried to split the fraction in four ones, multiply both the nominator and denominator by a power of nominator necessary to get rid of the cubic root, like:
$$ \frac{x + h}{h(x + h)^{1/3}} $$
But this didn't help, since dangling 'h's were still remaining in the expressions and I needed to combine the four fractions into a single one again, thus multiplying the nominators by cubic roots again.
Tried approach 2:
This one actually worked, and I'll post it as an answer.
Let's split the limit fraction in two ones:
$$ \lim_{h \to 0} \frac{(x + h)^{2/3} - (x - 1 + h)^{1/3} - x^{2/3} + (x- 1)^{1/3}}{h} = \\ \lim_{h \to 0} (\frac{(x + h)^{2/3} - x^{2/3}}{h} - \frac{(x- 1)^{1/3} - (x - 1 + h)^{1/3} }{h}) $$
For simplicity let's deal with the fractions without the limit notation:
$$ \frac{(x + h)^{2/3}- x^{2/3}}{h} \text{let's call it the left part} - \frac{(x- 1)^{1/3} - (x - 1 + h)^{1/3}}{h} \text{let's call it the right part} $$
I multiplied both the left and right parts to make the nominators become differences of the cubes (to get rid of the cubic roots).
The left part turned to be this:
$$ \require{cancel} \frac{(x + h)^2 - x^2}{h((x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3})} = \\ \frac{\cancel{x^2} + 2\cancel{h}x + \cancelto{h}{h^2} - \cancel{x^2}}{\cancel{h}((x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3})} = \\ \frac{2x + h}{(x + h)^{4/3} + x^{2/3}(x + h)^{2/3} + x^{4/3}} $$
the right part:
$$ \require{cancel} \frac{\cancel{x} - \cancel{1} - \cancel{x} + \cancel{1} - \cancelto{1}{h}}{\cancel{h}((x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + h)^{1/3} + (x - 1 + h)^{2/3})} = \\ \frac{- 1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + h)^{1/3} + (x - 1 + h)^{2/3}} $$
Since the h tends to 0, the limit is:
$$ \require{cancel} \lim_{h \to 0}(\frac{2x + \cancelto{0}{h}}{(x + \cancelto{0}{h})^{4/3} + x^{2/3}(x + \cancelto{0}{h})^{2/3} + x^{4/3}} - \frac{1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1 + \cancelto{0}{h})^{1/3} + (x - 1 + \cancelto{0}{h})^{2/3}}) = \\ \frac{2x}{x^{4/3} + x^{2/3}x^{2/3} + x^{4/3}} - \frac{1}{(x - 1)^{2/3} + (x - 1)^{1/3}(x - 1)^{1/3} + (x - 1)^{2/3}} = \\ \frac{2\cancel{x}}{3\cancelto{x^{1/3}}{x^{4/3}}} - \frac{1}{3(x - 1)^{2/3}} = \\ \frac{2}{3x^{1/3}} - \frac{1}{3(x - 1)^{2/3}} $$