# Tangent line

1. Dec 7, 2007

### fitz_calc

1. The problem statement, all variables and given/known data

if the tangent line to the graph of y=f(x) at (2.3) has an equation x-y+1=0, then f'(2) =?

3. The attempt at a solution

We did these problems in one small session at the beginning of the semester, but my notes aren't clear and I am not sure where to begin. Do I just take the first derivative of the equation and then find f'(2)?

2. Dec 7, 2007

### arunbg

If y=f(x) is a curve, then geometrically what is f'(x) ?
Can you solve now?

3. Dec 7, 2007

### HallsofIvy

Staff Emeritus
As I said on your other post (which was going the other way, from function to tangent), the derivative of a function, at a given x, is the slope of the tangent line to the graph there.

4. Dec 7, 2007

### fitz_calc

y=x+1
y=1+0
y=1

This is correct, but is my work correct -- or is the answer right only by coincidence?

5. Dec 7, 2007

### arunbg

Not correct. Did you follow the previous posts? y'(x) is the tangent of the curve y(x) for all valid x. You are given the equation of the tangent directly, and you are asked to find y'(2). Can you proceed?

6. Dec 7, 2007

### Office_Shredder

Staff Emeritus
Well, it is technically a correct method, as the slope of the line y=x+1 is found by taking the derivative

7. Dec 7, 2007

### arunbg

Hmm, according to the question the slope at point (2,3) is already given by y=x+1(correct me if my interpretation is wrong).

8. Dec 7, 2007

### HallsofIvy

Staff Emeritus
What do you mean by "given by"? We are told that the tangent line is y= x+ 1. The derivative of the curve is the slope of the tangent line, which is 1.

9. Dec 7, 2007

### arunbg

Oops bad quick post, slope should be changed to tangent. Thanks for pointing out Halls. It's a matter of simple substitution of course. No differentiation involved.