# Thread: Having trouble with basic Derivatives

1. ## Having trouble with basic Derivatives

Allright, I went to the lecture, read the section in my math book, and watched videos on youtube..., but still...how to arrive to the answer to this problem eludes me.

Problem:

Find an equation of the tangent line to the curve at the given point.

y = (x-1)/(x-2), (3,2)

Equation that the book gives to solve this problem:

M = lim x -> a [f(x) - f(a)] / [x - a] where point (a, f(a))

My Work:

f(x) = (x-1)/(x-2)
f(a) = 2
a = 3

---

M= [[(x-1)/(x-2)] - 2] / [x-3]

---

M= [(x-1)/(x-2) - (2x-4)/(x-2)] / (x-3)

---

M= [(-x+3)/(x-2)] / (x-3)

---

M= (-x+3) / [(x-2)(x-3)]

~This is as far as I get. I have no idea if I am correct so far. All I know is that I am not getting the answer that the book gives in the back~

Answer the book gives: y = -x+5

2. You need to remember to take the $\displaystyle \lim_{x \to a}$ part down each time...

Assuming you had done this you would have gotten to

$\displaystyle M= \lim_{x \to 3}\frac{(-x + 3)}{(x - 2)(x - 3)}$

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

$\displaystyle = \lim_{x \to 3}\frac{-1}{x - 2}$

$\displaystyle = \frac{-1}{3 - 2}$

$\displaystyle = -1$.

So that means for your tangent line, which is of the form $\displaystyle y = Mx + c$, that $\displaystyle M = -1$.

So you have $\displaystyle y = -x + c$.

Use the point that lies on the curve to find the value of $\displaystyle c$.