Wouldn’t it be cool if you could use our derivative rules rather than using the limit definition of derivative?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
Great question, and we’re going to answer it in today’s calculus class.
Let’s go!
Background
Now, I know what you may be thinking, why does this matter if we already know how to evaluate a limit?
Because there are times when evaluating the limit definition of derivative is tedious and downright challenging to do algebraically.
Besides the fact, if we can recognize that a limit represents the derivative of some function, then wouldn’t it be easier to cut to the chase and take the derivative using our known derivative laws?
Yes! Yes, it would!
The Formula
Okay, so let’s learn the strategies and tricks for finding the derivative quickly by spotting a derivative definition!
So, if we can spot f(x) and recognize that we are being asked to find the instantaneous rate of change, then we can use our derivative rules.
How to calculate the slopes of tangents and velocities
Example #1
Let’s put this idea to the test with a few examples.
\begin{equation}
\text { Find } \lim _{h \rightarrow 0} \frac{(x+h)^{2}-x^{2}}{h}
\end{equation}
First, let’s see if we can spot f(x) from our limit definition of derivative.
\begin{equation}
\lim _{h \rightarrow 0} \frac{(x+h)^{2}-x^{2}}{h} \Leftrightarrow \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\end{equation}
This means what we are really being asked to find is \(f^{\prime}(x)\) when \(f(x)=x^{2}\).
\begin{equation}
\lim _{h \rightarrow 0} \frac{(x+h)^{2}-x^{2}}{h}=f^{\prime}(x) \text { where } f(x)=x^{2}
\end{equation}
So, rather than use the limit definition of derivative, let’s use the power rule instead!
\begin{equation}
\begin{array}{l}
f^{\prime}(x)=2 x \\
\lim _{h \rightarrow 0} \frac{(x+h)^{2}-x^{2}}{h}=2 x
\end{array}
\end{equation}
Cool, right?
Example #2
Alright, so now let’s suppose we are given the following limit, and we are asked to evaluate.
\begin{equation}
\lim _{x \rightarrow 1} \frac{x^{9}-1}{x-1}
\end{equation}
Well, when we substitute 1 in for every x, we get 0/0 which is an indeterminate form.
So this means that we can factor, use L’Hospital’s Rule, or recognize we have been given a derivative in disguise!
\begin{equation}
\lim _{x \rightarrow 1} \frac{x^{9}-1}{x-1} \Leftrightarrow \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}
\end{equation}
This means we are being asked to find the derivative of the function f(x) at a number a!
\begin{equation}
\lim _{x \rightarrow 1} \frac{x^{9}-1}{x-1}=f^{\prime}(1) \text { where } f(x)=x^{9}
\end{equation}
So, once again, rather than use the limit definition of derivative, let’s use the power rule and plug in x = 1 to find the slope of the tangent line.
\begin{equation}
\begin{array}{l}
f^{\prime}(x)=9 x^{8} \\
f^{\prime}(1)=9(1)^{8}=9 \\
\lim _{x \rightarrow 1} \frac{x^{9}-1}{x-1}=9
\end{array}
\end{equation}
Amazing!
Summary
And for all you high school calculus students, you’re bound to get a question like this on the AP Calculus Exam, so let’s make sure we know what to do.
Together we will learn how to quickly recognize the definition of the derivative and then use our derivative rules to arrive at our final answer swiftly and efficiently.
Video Tutorial w/ Full Lesson & Detailed Examples (Video)
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