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I am having trouble with understanding the following question:

Find the derivative of the given function: $f(x)=\frac{x}{2x+\frac{1}{3x+1}}$

I have always thought that in order for a function to have a derivative, it must be continuous. This function is clearly undefined for $f(-\frac{1}{3})$. So, how is it possible for it to have a derivative?

Ignoring that, I have tried to solve for the derivative but my textbook is disagreeing with me on the answer. Here's what I did

$f(x)=\frac{x}{2x+\frac{1}{3x+1}}$

$f(x)=\frac{x}{\frac{6x^2+2x+1}{3x+1}}$

$f(x)=\frac{x}{1}\cdot\frac{3x+1}{6x^2+2x+1}$

$f(x) =\frac{x}{1}\cdot\frac{3x+1}{2x(3x+1)+1}$

$f(x) =\frac{x}{1}\cdot\frac{1}{2x+1}$

$f(x) =\frac{x}{2x+1}$

$f'(x) =\frac{(2x+1)(x)'-(x)(2x+1)'}{(2x+1)^2}$

$f'(x) =\frac{1}{(2x+1)^2}$

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    It's good to check your algebra by plugging in a value from time to time, especially when you're getting an answer which doesn't seem to work. Try $x=1$ in your first few steps and see if you can find where the problem has to be.2012-10-06

3 Answers 3

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The problem is with your algebra: $\frac{3x+1}{2x(3x+1)+1}\ne\frac1{2x+1}\;,$ as you can check by verifying that $(3x+1)(2x+1)\ne 2x(3x+1)+1$.

It’s true that $f$ isn’t defined everywhere and therefore isn’t differentiable everywhere, but it turns out to be differentiable almost everywhere, and the derivative is the same function everywhere that it’s defined.

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HINT: $\frac{a+b}{c(a+b)+d}\neq\frac{1}{c+d}$

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    I realized that about 5 minutes after posting it. *facepalm*2012-10-06
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The function has a derivative everywhere but at $-2/3$, that's okay. Your mistake is between step 4 and 5, you can't pull $3x + 1$ out like that.