# Math Help - Derivative, where am I going wrong in this problem?

1. ## Derivative, where am I going wrong in this problem?

We need to find the equation of a tangent line, and to do that you need to find derivative to be able to complete the equation.

So, it gives y= (x-1)/(x-2) on coordinate (3,2)

I say:
a=3
f(a)=2
f(3)= ?

Using lim h-->0 f(a+h)-f(a) / h to find f(3)

I plug in [(3+h-1)/(3+h-2) -2] / h

=(3+h-1)-2(3+h-2) / h(3+h-2) (After multiplying by CD)
=3+h-1-6-2h+4 / 3h+h^2-2h
=-2 / 2h+h-2
=-2/-2
=1

The book is showing the derivative as -1. Where did I mess up here?

2. Originally Posted by Marconis

I plug in [(3+h-1)/(3+h-2) -2] / h
This is not complete

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

Then sub x=3

3. $\displaystyle f'(3) = \lim_{x \to 3} \frac{f(x)-f(3)}{x-3}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{\frac{x-1}{x-2} - 2}{x-3}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{(x-1) - 2(x-2)}{(x-3)(x-2)}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{3-x}{(x-3)(x-2)}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{-1}{x-2} = -1$

4. Originally Posted by skeeter
$\displaystyle f'(3) = \lim_{x \to 3} \frac{f(x)-f(3)}{x-3}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{\frac{x-1}{x-2} - 2}{x-3}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{(x-1) - 2(x-2)}{(x-3)(x-2)}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{3-x}{(x-3)(x-2)}$

$\displaystyle f'(3) = \lim_{x \to 3} \frac{-1}{x-2} = -1$
That is how the book shows it, but my professor taught us using the formula that I stated in my original post. Despite this, she informed us that they are the same exact things. So, doing it my way, where am I going wrong?

5. Originally Posted by Marconis
We need to find the equation of a tangent line, and to do that you need to find derivative to be able to complete the equation.

So, it gives y= (x-1)/(x-2) on coordinate (3,2)

I say:
a=3
f(a)=2
f(3)= ?

Using lim h-->0 f(a+h)-f(a) / h to find f(3)

I plug in [(3+h-1)/(3+h-2) -2] / h

=(3+h-1)-2(3+h-2) / h(3+h-2) (After multiplying by CD)
=3+h-1-6-2h+4 / 3h+h^2-2h
= -h/[h(h+1)]
= -1/(h+1)
= -1

6. EDIT

Got it! Thanks. I wrote it more neatly and I was able to get it, was also stupidly distributing incorrectly. Thanks again.

7. Originally Posted by Marconis
We need to find the equation of a tangent line, and to do that you need to find derivative to be able to complete the equation.

So, it gives y= (x-1)/(x-2) on coordinate (3,2)

I say:
a=3
f(a)=2
f(3)= ?

Using lim h-->0 f(a+h)-f(a) / h to find f(3)

I plug in [(3+h-1)/(3+h-2) -2] / h

=(3+h-1)-2(3+h-2) / h(3+h-2) (After multiplying by CD)
=3+h-1-6-2h+4 / 3h+h^2-2h
In the numerator, 3- 1- 6+ 4= 0 and h- 2h= -h. In the denominator 3h+ h^2- 2h= h+ h^2= h(1+ h)

=-2 / 2h+h-2
so I don't see how you get this. It should be -1/(1+ h).

=-2/-2
=1

The book is showing the derivative as -1. Where did I mess up here?