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Derivative Using the Limit Definition

Now that we have defined limits and are able to find them, we can begin to talk about the second major topic of Calculus, derivatives. In this example, we will show how to calculate these derivatives using the formal limit definition. If you would like to learn more about the motivation for such a definition and how to derive it, I hope that you will visit my post Dividing by zero: Imploding the Universe. I also have a discovery learning exercise available on my products page, Pizza and Marginal Profit that uses marginal profit at a pizzeria to motivate finding a derivative.

Find the derivative

For this example we will use the limit definition of derivative to show that
\begin{align*}
\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}.
\end{align*}
Before we start our calculations, we will remind ourselves that the derivative of a function is
\begin{align*}
\frac{d}{dx}f(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}.
\end{align*}

Then, we will proceed to find this one step at a time.

\(1. f(x+h)\)

We begin by finding \(f(x+h)\). Since \(f(x)=\sqrt{x}\), we find that
\begin{align*}
f(x+h)&=\sqrt{x+h}.
\end{align*}
If we simplify this at all, we would. However, there isn’t anything else we can do right now.

\(2. f(x+h)-f(x)\)

We can now find
\begin{align*}
f(x+h)-f(x)&=\sqrt{x+h}-\sqrt{x}.
\end{align*}

While it doesn’t look like we can simplify this, we note that eventually we will have to take a limit. As we saw in Algebraic limits, if we have a square root in our limit, we will need to multiply by the conjugate in order to simplify the problem. Instead of waiting until later, we will do this now.

\begin{align*}
f(x+h)-f(x)&=\sqrt{x+h}-\sqrt{x} \\
&=\left(\sqrt{x+h}-\sqrt{x}\right)\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right) \\
&=\frac{x+h-x}{\sqrt{x+h}+\sqrt{x}} \\
&=\frac{h}{\sqrt{x+h}+\sqrt{x}}.
\end{align*}

We have now simplified this portion as much as we can.

\(3. \frac{f(x+h)-f(x)}{h}\)

We now find the quotient of our previous term with \(h\) and get that
\begin{align*}
\frac{f(x+h)-f(x)}{h}&=\frac{\left(\frac{h}{\sqrt{x+h}+\sqrt{x}}\right)}{h} \\
&=\left(\frac{h}{\sqrt{x+h}+\sqrt{x}}\right)\frac{1}{h} \\
&=\frac{1}{\sqrt{x+h}+\sqrt{x}} \text{ if } h \neq 0.
\end{align*}
Note that in the second to last line we divided fractions by multiplying by the reciprocal. We can then cancel \(\frac{h}{h}=1\) as long as \(h \neq 0\). Since we will be taking the limit as \(h\) goes to \(0\), we won’t have to worry about the case when \(h=0\).

\(4. \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}\)

We are finally able to finish finding our derivative.

\begin{align*}
\frac{d}{dx}\sqrt{x}&=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \\
&=\lim_{h \to 0}\frac{1}{\sqrt{x+h}+\sqrt{x}} \\
&=\frac{1}{\sqrt{x+0}+\sqrt{x}} \\
&=\frac{1}{2\sqrt{x}}.
\end{align*}
Hence, we have shown exactly what we set out to prove, that is, \(\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\).

Conclusion

Here we were able to find the derivative of \(\sqrt{x}\) using the limit definition of derivative. While it is possible to do this all in one step, I highly suggest that you don’t do this. In general, the more things you try to do at once, the more likely you will make an error. As such, I use the four step process outlined above to find derivatives with limits. If you’d like to see more on this, I first learned this method from Karl Byleen while at Marquette. His Calculus book uses this method, and it is worth trying even if you are used to finding derivatives.

I do also want to note that for this problem I gave the solution instead of making you find it. This is to highlight that, in general, the purpose of questions like this is to ensure that you understand what a derivative is, not that you can just find it using a derivative rule. Later in Calculus, finding derivatives will become much quicker, but knowing the definition in terms of limits will remain important.

I hope you learned something and that this helped with you studying. If it did, make sure to like the post and share it on Social Media with anyone else that may need some help with derivatives.

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