Derivatives. If you’re currently taking Calc 1 (which you probably are if you found yourself here), you are probably up to your elbows in derivative problems. One type is taking the derivative of a fraction, or better put, a *quotient*. That’s what this post is about. There is nothing special about this situation, I just needed a post for this month and this search term came up in Google Suggest so I figured, why not?\( \)

To find the derivative of a fraction, you use the quotient rule:

\begin{equation*}

\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x) – f(x)g'(x)}{[g(x)]^2}

\end{equation*}

That’s it. It’s the best case scenario in math: just plug into the formula. You don’t have to understand where the formula came from, you just have to remember it. Some people remember it this way:

\begin{equation*}

\frac{\text{LoDHi – HiDLo}}{\text{Lo}^2}

\end{equation*}

Frankly, I don’t find this very helpful, as I get the “Lo’s” and the “Hi’s” mixed up. I just remember that the denominator comes first on top. You just find a way that works for you and go with it. Now for some examples:

##### Example 1

Find the derivative:

\begin{equation*}

h(x) = \frac{2}{x+1}

\end{equation*}

If you’re worried about putting everything in the right place in the formula, it may help to write out \(f(x)\) and \(g(x)\) separately, as well as their derivatives:

\begin{array}{cc}

f(x) = 2 & g(x) = x+1\\

f'(x) = 0 & g'(x) = 1

\end{array}

Now just plug everything in:

\begin{equation*}

h'(x) = \frac{(x+1)\cdot 0 – 2\cdot 1}{(x+1)^2} = \frac{-2}{(x+1)^2}

\end{equation*}

Hooray! (Note: An alternative method would be to write the function as \(h(x) = 2(x+1)^{-1}\) and use the power and chain rules. This is actually how I would do this particular problem, as I try to avoid the quotient rule at all costs. But that’s just me.)

##### Example 2

Find the derivative:

\begin{equation*}

h(x) = \frac{x^3-4x}{5x^2+x+1}

\end{equation*}

This is the same as the last example, only with slightly more complicated expressions. Start by assigning \(f(x) = x^3-4x\) and \(g(x) = 5x^2+x+1\). Then we have

\begin{array}{cc}

f(x) = x^3-4x & g(x) = 5x^2+x+1\\

f'(x) = 3x^2-4 & g'(x) = 10x+1

\end{array}

Now just plug everything in.

\begin{equation*}

h'(x) = \frac{(5x^2+x+1)(3x^2-4) – (x^3-4x)(10x+1)}{(5x^2+x+1)^2}

\end{equation*}

You’re not done. No (decent) calculus teacher will let you get away with leaving your answer like this. You have to simplify it. **Make sure you use parentheses in the numerator.** Otherwise, you *will* mess up with that minus sign. After multiplying the numerator out and collecting like terms, you should get

\begin{equation*}

h'(x) = \frac{5x^4 + 2x^3 + 23x^2 – 4}{(5x^2+x+1)^2}

\end{equation*}

Now the next thing you have to ask yourself is: Does the numerator have a factor of \(5x^2 + x + 1\)? This is because if it does, you can simplify it further by canceling a factor in the denominator. You can figure this out by using polynomial division. I’m not going to do that here, though. Most teachers would be ok with you just leaving like this.

##### Example 3

Find the derivative:

\begin{equation*}

h(x) = \frac{\sin x}{1 + \cos x}

\end{equation*}

This problem is a good example of using trig identities. I’m going to just going to plug straight into the formula this time:

\begin{equation*}

h'(x) = \frac{(1+\cos x)D\{\sin x\} – (\sin x)D\{1 + \cos x\}}{(1+\cos x)^2} =

\frac{(1 + \cos x)(\cos x) – (\sin x)(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}

\end{equation*}

Now, just because you multiplied the numerator out doesn’t mean the thing is completely simplified. When dealing with trig functions, you always have to check if there are any identities you can apply. In this case, we can use everyone’s favorite identity, which is \(\sin^2 x + \cos^2 x = 1\). We get

\begin{equation*}

h'(x) = \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2} = \frac{1 + \cos x}{(1+\cos x)^2} = \frac{1}{1+ \cos x}

\end{equation*}

Isn’t that neat how we were able to cancel a factor out of the denominator? I love it when that happens :).

##### Example 4

Find the derivative:

\begin{equation*}

h(x) = \frac{\sqrt{\ln x}}{x}

\end{equation*}

This is a problem where you have to use the chain rule. My advice for this problem is to find the derivative of the numerator separately first. So if \(f(x) = \sqrt{\ln x}\), we can write \(f(x) = (\ln x)^{1/2}\), so

\begin{equation*}

f'(x) = \frac{1}{2}(\ln x)^{-1/2}\frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}

\end{equation*}

Now, let’s find the derivative of \(h(x)\). Plugging straight into the formula, we get

\begin{equation*}

h'(x) = \frac{x\frac{1}{2x\sqrt{\ln x}} – \sqrt{\ln x}}{x^2}

\end{equation*}

Now, because it is so complicated, you might be tempted to just leave it like this. But you shouldn’t. This needs to be simplified. Start with the obvious: cancel the \(x\) in the first term in the numerator. Next, put the terms in the numerator over a common denominator, which is \(2\sqrt{\ln x}\)

\begin{equation*}

h'(x) = \frac{x\frac{1}{2x\sqrt{\ln x}} – \sqrt{\ln x}}{x^2} = \frac{\frac{1}{2\sqrt{\ln x}} – \sqrt{\ln x}}{x^2} = \frac{\frac{1}{2\sqrt{\ln x}} – \frac{2\ln x}{2 \sqrt{\ln x}}}{x^2} = \frac{1-2\ln x}{2x^2 \sqrt{\ln x}}

\end{equation*}

Fun, huh? When you’re doing these kinds of problems, just remember: it’s making you smarter.

Hopefully, these examples give you some ideas for how to find the derivative of a fraction. If you have any comments or questions, please leave them below!