Typical Calculus Exam Problem: Definition of the Derivative

This week’s problem is a typical exam question that requires using the Definition of the Derivative to first find the value of a function’s derivative at a given point, and then use that value to write the equation for the tangent line to the curve at that point.

Recall the primary form of the Definition of the Derivative:

A problem similar to this one is often on the first midterm in Calculus 1:

Solution.

(a) We apply the definition of the derivative to the given function, $f(x) = \sqrt{x+3}:$ \begin{align*} f'(x) &= \lim_{h \to 0}\frac{f(x+h) – f(x)}{h} \\[8px] &= \lim_{h \to 0}\frac{\sqrt{(x+ h) + 3} – \sqrt{x + 3}}{h} \hphantom{ \cdot \frac{\sqrt{x+ h + 3} + \sqrt{x + 3}}{\sqrt{x+ h + 3} + \sqrt{x + 3}}} \end{align*} This is now a straightforward “find the limit” problem. We use conjugates as we’ve been practicing: \begin{align*} &= \lim_{h \to 0}\frac{\sqrt{x+ h + 3} – \sqrt{x + 3}}{h} \cdot \frac{\sqrt{x+ h + 3} + \sqrt{x + 3}}{\sqrt{x+ h + 3} + \sqrt{x + 3}} \\[8px] &= \lim_{h \to 0} \frac{(x+h+3) – (x+ 3)}{h \Big(\sqrt{x+ h + 3} + \sqrt{x + 3} \Big)} \\[8px] &= \lim_{h \to 0} \frac{\cancel{h}}{\cancel{h} \Big(\sqrt{x+ h + 3} + \sqrt{x + 3} \Big)} \\[8px] &= \lim_{h \to 0} \frac{1}{\sqrt{x+ h + 3} + \sqrt{x + 3}} \\[8px] &= \frac{1}{\sqrt{x + 3} + \sqrt{x + 3}} \\[8px] &= \frac{1}{2\sqrt{x + 3}} \end{align*}

Then \[f'(1) = \frac{1}{2\sqrt{1 + 3}} = \frac{1}{4} \quad \cmark\]

(b) We use our standard strategy to write the equation of a tangent line, which uses the point-slope form of a line: \[\textbf{Point-Slope Form of a Line:} \quad y – y_0 = m \left(x – x_0 \right) \] (Don’t remember that equation? It’s just the definition of slope, rewritten: $m = \dfrac{y – y_0}{x – x_0}$.)

We want the tangent line at $x_0 = 1.$ We found the slope of that tangent line in part (a): \[m = f'(1) = \frac{1}{4} \;\blacktriangleleft\] We need the y-value of the function $f(x) = \sqrt{x+3}$ at $x_0 = 1$: \[y_0 = f(x_0) = \sqrt{1+3} = 2 \; \blacktriangleleft \] We can then easily write the equation for the tangent line with $m = \dfrac{1}{4}$ and $\left(x_0, y_0 \right) = (1, 2):$ \[y – 2 = \frac{1}{4}\left( x – 1\right) \quad \cmark\]

The graph shows the curve $y = f(x)$ and the tangent line as requested.

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