# Thread: Using quotient and chain rules for derivatives

1. ## Using quotient and chain rules for derivatives

Hey so I have a calculus assignment and im not wrapping my head around many of the concepts for derivatives.

My quotient rule question:

a) Use the definition of the derivative to find f'(x) for f(x) = (1/3x-4).
b) Use the result from a) to fi nd the equation of the tangent line to f(x) at the point (2, 1/2 ).

For this I believe I have answered the first part, with

f(x)=(1/3x-4)

f'(x)=(f(x)*g'(x)-f'(x)g(x))/f(x)^2

=-3/(3x^2-24x+16)

I'm don't know what to do to continue to part be.
Thanks very much for any help you might be able to offer.

2. Originally Posted by commerce-calc

=$\displaystyle \frac{-3}{3x^2-24x+16}$

I'm don't know what to do to continue to part be.
Thanks very much for any help you might be able to offer.
Substitute the Point 2 into $\displaystyle \frac{-3}{3x^2-24x+16}$ , which will give you the gradient of the tangent line.

Now, for y-coordinate, subtitute the point x=2 into f(x) i.e. $\displaystyle \frac{1}{3x-4}$, which will give you the y-coordinate.

Then simply put the data into the either 2 formulas Point-Slope or Slope-Intercept, I recommend point-slope.

and there you go, hope you understood

3. awesome. that actually makes sense im not used to that haha.

so i came up with the equation for the tangent line:

y=3/10x-01

sound maybe right? thanks for clearing that up for me

4. Happy to help, just going along these things

5. Originally Posted by commerce-calc

=-3/(3x^2-24x+16).
Just noticed this, 3x^2 shouldn't be 3x, it should be 9x^2, correct it, and make the equation again, this time you'll get the right one, sorry for not pointing out earlier

6. haha no problem. and thank you very much for noticing the error, fixed and good. thanks again for the help.