TITLE: derivative of $\left(\frac{\ 1}{3x}\right)$ using limit definition.
QUESTION [7 upvotes]: I'm trying find derivative of $\left(\frac{\ 1}{3x}\right)$ using limit definition.
I do this:
$(\left(\frac{\ 1}{3x+h}\right) - \left(\frac{\ 1}{3x}\right))/h$
$(\left(\frac{\ 1}{3x+h}\right)(3x) - \left(\frac{\ 1}{3x}\right)(3x+h))/h$
$(\left(\frac{\ 3x}{3x(3x+h)}\right) - \left(\frac{\ 3x+h}{3x(3x+h)}\right))/h$
$\left(\frac{\ 3x - 3x - h}{3x(3x+h)}\right)/h$
$\left(\frac{\  - h}{3x(3x+h)}\right)/h$
$\left(\frac{\  - h}{3x(3x+h)}\right)* \left(\frac{\   1}{h}\right)$
$\left(\frac{\  - 1}{3x(3x+h)}\right)$
Answer: $\left(\frac{\  - 1}{9x^2}\right)$
But in https://www.symbolab.com/solver/derivative-calculator
the answer is $\left(\frac{\  - 1}{3x^2}\right)$
I think my calculations are correct. How can symbolad answer be different?
Thanks!

REPLY [11 votes]: Your beginning step was off. You needed to write $\dfrac{1}{3(x+h)}$, not $\dfrac{1}{3x+h}$. When you replace $x$ with $x+h$, you need to be careful about putting () around the $x+h$ (or at least imagining you are) so that you maintain the same function.

REPLY [10 votes]: I do this:
$(\left(\frac{\ 1}{3x+h}\right) - \left(\frac{\ 1}{3x}\right))/h$

Careful! If $f(x)=\tfrac{1}{3x}$, then:
$$f\left( \color{blue}{x+h} \right) = \frac{1}{3\left( \color{blue}{x+h} \right)} = \frac{1}{3x+3h} \color{red}{\ne} \frac{1}{3x+h}$$
Can you fix your work? The rest is fine!