Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 47

Suppose that $f(x) = \sqrt[3]{x}$, find $f'(x), f''(x), f'''(x)$ and $f^4(x)$. Graph $f, f', f''$ and $f'''$ on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?

a.) Let $a \neq 0$, use the definition of derivative $\displaystyle f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a}$ to find $f'(a)$.

Using the definition of derivative


$\begin{equation} \begin{aligned} \qquad f'(a) =& \lim_{x \to a} \frac{\sqrt[3]{x} - \sqrt[3]{a}}{x - a} \cdot \frac{\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}}{\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}} && \text{Multiply both numerator and denominator by $\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}$} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{x + \cancel{\sqrt[3]{ax^2}} + \cancel{\sqrt[3]{a^2x}} - \cancel{\sqrt[3]{ax^2}} + \cancel{\sqrt[3]{a^2x}} - a}{(x - a)(\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2})} && \text{Combine like terms} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{\cancel{x - a}}{\cancel{(x - a)}(\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}) } && \text{Cancel out like terms} \\ \\ \qquad f'(a) =& \lim_{x \to a} \left( \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}} \right) = \frac{1}{\sqrt[3]{a^2} + \sqrt[3]{(a)(a)} + \sqrt[3]{a^2}} && \text{Evaluate the limit} \\ \\ f'(a) =& \frac{1}{\sqrt[3]{a^2} + \sqrt[3]{a^2} + \sqrt[3]{a^2}} && \text{Combine like terms} \\ \\ f'(a) =& \frac{1}{3 \sqrt[3]{a^2}} \text{ or } \frac{1}{3(a)^{\frac{2}{3}}} && \end{aligned} \end{equation}$



b.) Prove that $f'(0)$ does not exist

Using $f'(a)$ in part (a)


$\begin{equation} \begin{aligned} f'(a) =& \frac{1}{3 \sqrt[3]{a^2}} \\ \\ f'(0) =& \frac{1}{3 \sqrt[3]{(0)^2}} \\ \\ f'(0) =& \frac{1}{3(0)} = \frac{1}{0} \end{aligned} \end{equation}$


Therefore, $f'(0)$ does not exist because denominator is zero.

c.) Prove that $y = \sqrt[3]{x}$ has a vertical tangent line at $(0,0)$

If the function has a vertical tangent line at $x = 0, \lim_{x \to 0} f'(x) = \infty$

Given that $f'(x) = \displaystyle \frac{1}{3 \sqrt[3]{x^2}}$

Suppose that we substitute a value closer to from left and right to the limit of $f'(x)$. Let's say $x = -0.00001$ and $x = 0.000001$


$\begin{equation} \begin{aligned} & \lim_{x \to 0^-} \frac{1}{3\sqrt[3]{(-0.00001)^2}} = 2154.43 \\ \\ & \lim_{x \to 0^+} \frac{1}{3 \sqrt[3]{(0.0000001)^2}} = 46415.89 \end{aligned} \end{equation}$


This means that where $x$ gets closer and closer to , the value of limit approached a very large number. The tangent line with these values become steeper and steeper as $x \to 0$ until such time that the tangent line becomes a vertical line at $x = 0$.