# Exponents On Slopes.

1. Sep 13, 2005

### Line

Reading on Calculas you can see that the slope of a tangent can have an exponet,but why?

In Alegra and Trig we were always taught that slope was a fraction and x was never sqaured or cubed like in The Linear Eqaution Y=MX+B.

In calculas DY can eqaul Xsqaured, Xcubed or to any degree. But how?

if you wrote Y=4/3X(sqaured)+6 you wouldn't get a line but some sort of curved shape. So why is it the tagent of a curve can have X to an exponent in Calculas?

2. Sep 13, 2005

### Galileo

What do you mean? The tangent at a certain point on a curve is a line by definition (and of the form y=mx+b). When you talk about the slope of a tangent you are talking about the slope of a line, which is constant.

Maybe you'd better give an example for us too see where your confusion lies.

3. Sep 13, 2005

### HallsofIvy

Staff Emeritus
The slope of a tangent line at some point on a curve is a number just like the slope of any line.

You may be confusing the slope of the tangent at a point with the "slope function" (the derivative) which gives the slope at different points.

For example, if f(x)= x3, then f'= 3x2- but that is not the "slope" at any specific point.
If x= 1, f'(1)= 3 and the tangent line at (1,1) is y= 3(x-1)+ 1= 3x- 2.
If x= 2, f'(2)= 24 and the tangent line at (2, 8)is y= 24(x-2)+ 8= 24x- 40.

4. Sep 13, 2005

### Line

OK how did you get 3X-2?

5. Sep 13, 2005

### VietDao29

A tangent line which goes through (x0, y0), can be expressed as:
$$y = f'(x_0) (x - x_0) + y_0$$
f(x) = x3
f'(x) = 3x2
So say x0 = 1, then y0 = f(x0) = 1.
The tangent line at (1, 1) is:
y = f'(x0) (x - x0) + y0 = f'(1) (x - 1) + 1 = 3x - 3 + 1 = 3x + 2.
-------------
There seems to be a slight typo with f'(2) , f'(2) = 12, not 24, so:
y = 12(x - 2) + 8 = 12x - 16.
Viet Dao,

Last edited: Sep 13, 2005
6. Sep 13, 2005

### HallsofIvy

Staff Emeritus
Oops! In calculating f'(2), I cubed 2 instead of squaring!

(and, just to get back at you- 3(x-1)+ 1= 3x-3+1= 3x- 2 not 3x+2!)

Last edited: Sep 13, 2005
7. Sep 13, 2005

### lurflurf

It is because x^3 or whatever is just a number. The idea is we like simple functions like y=m*x+b so when confronted by a (differentiable) function f(x) we consider that in the limit dx->0 the function is linear near some point a
f(x)=f(a)+f'(a)x+O(x^2)
so f'(a) is just the slope be it a or a^3 or exp(a)
the linearity is in another variable.

8. Sep 13, 2005

### Line

SO is it 3x-2 or 3x+2?

9. Sep 16, 2005

### VietDao29

Yup. Thanks for pointing that out...
It's 3x - 2.
------------------------
I'll make it a bit clear for you.
Let f(x) be some function.
f'(x) is the derivative of f(x).
Then f'(x) is the function such that when you plug any x0 into the function, it will return you the slope of the tangent line at the point (x0, f(x0)).
Say f(x) = 4x3. So f'(x) = 12x2. If you want to have the slope of the tangent line at (1, 4) (f(1) = 4 . 13 = 4), you plug x = 1 to f'(x). It will return f'(1) = 12.
So the slope of the tangent line at (1, 4) is 12.
------------------------
And if you want to find out the equation of the tangent line at (1, 4), you use the equation: y = f'(x0) (x - x0) + y0. Where y0 = f(x0). I think you can find the proof in your book.
So the tangent line at (1, 4) is y = f'(1) (x - 1) + 4 = 12 (x - 1) + 4 = 12x - 12 + 4 = 12x - 8. So y = 12x - 8.
Viet Dao,