Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 30

a.) Suppose that $f(t) = t^2 - \sqrt{t}$, find $f'(t)$.

Using the definition of derivative




$\begin{equation} \begin{aligned} \qquad f'(t) =& \lim_{h \to 0} \frac{f(t + h) - f(t)}{h} && \\ \\ \qquad f'(t) =& \lim_{h \to 0} \frac{(t + h)^2 - \sqrt{t + h} - (t^2 - \sqrt{t})}{h} && \text{Substitute $f(t + h)$ and $f(t)$} \\ \\ \qquad f'(t) =& \lim_{h \to 0} \frac{\cancel{t^2} + 2th + h^2 - \sqrt{t + h} - \cancel{t^2} + \sqrt{t}}{h} && \text{Expand the equation and combine like terms} \\ \\ \qquad f'(t) =& \lim_{h \to 0} \frac{2th +h^2 - \sqrt{t + h} + \sqrt{t}}{h} && \text{Isolate the terms that has square root and multiply it by its conjugate} \\ \\ \qquad f'(t) =& \lim_{h \to 0} \frac{2th + h^2}{h} + \frac{\sqrt{t} - \sqrt{t + h}}{h} \cdot \frac{\sqrt{t} + \sqrt{t + h}}{\sqrt{t} + \sqrt{t + h}} && \text{Get the factor of the first term and multiply both numerator and denominator of the second term by $(\sqrt{t} + \sqrt{t + h} )$} \\ \\ \qquad f'(t) =& \lim_{h \to 0} \frac{\cancel{h}(2t + h)}{\cancel{h}} + \frac{t - \cancel{\sqrt{t(t + h)}} + \cancel{\sqrt{t (t + h)}} - (t + h)}{h(\sqrt{t} + \sqrt{t + h})} && \text{Cancel out and combine like terms} \\ \\ \qquad f'(t) =& \lim_{h \to 0} 2t + h + \frac{\cancel{t} - \cancel{t} - h}{h(\sqrt{t} + \sqrt{t + h})} && \text{Combine like terms in the second term} && \\ \\ \qquad f'(t) =& \lim_{h \to 0} 2t + h - \frac{\cancel{h}}{\cancel{h}(\sqrt{t} + \sqrt{t + h})} && \text{Cancel out like terms in the second term} \\ \\ \qquad f'(t) =& \lim_{h \to 0} \left( 2t + h - \frac{1}{\sqrt{t} + \sqrt{t + h}} \right) = 2t + 0 - \frac{1}{\sqrt{t} + \sqrt{t + 0}} = 2t- \frac{1}{\sqrt{t} + \sqrt{t}} && \text{Evaluate the limit} \\ \\ f'(t) =& 2t - \frac{1}{2 \sqrt{t}} && \end{aligned} \end{equation}$


b.) Compare the graphs of $f$ and $f'$ and check whether your answer in part (a) is reasonable.