7

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!

user21820
  • 57,693
  • 9
  • 98
  • 256
Cassio Ribeiro
  • 71
  • 4
    you must write $$\frac{1}{3(x+h)}$$ – Dr. Sonnhard Graubner Apr 10 '17 at 12:08
  • 2
    As a side note your second line is incorrect. You need to multiply the first term by (3x)/(3x) but you only show multiplying by 3x. Similarly for the second term you should be using (3x+h)/(3x+h) instead of just 3x+h (although now you know that 3(x+h) is the correct thing anyways). Your next line treats it properly so it might have just been an issue with writing down each step here but be careful about those kinds of things. – Dason Apr 10 '17 at 12:34

4 Answers4

11

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.

Will Craig
  • 845
10

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!

Community
  • 1
StackTD
  • 27,903
  • 34
  • 63
1

$$\lim _{h\to 0}\left(\frac{\frac{1}{\color{red}{3\left(x+h\right)}}-\frac{1}{3x}}{h}\right)=\lim _{h\to \:0}\left(-\frac{1}{3x\left(x+h\right)}\right)=-\frac{1}{3x^2}$$

Amarildo
  • 2,162
  • 14
  • 20
1

The value of $x \mapsto 3x$ at $x+h$ is $3(x+h)=3x+3h$ not $3x+h$.

Olivier Oloa
  • 120,989