# Exponents On Slopes.

#### 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?

#### Galileo

Science Advisor
Homework Helper
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.

#### HallsofIvy

Science Advisor
Homework Helper
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.

#### Line

OK how did you get 3X-2?

#### VietDao29

Homework Helper
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.
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HallsofIvy said:
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.
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:

#### HallsofIvy

Science Advisor
Homework Helper
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 by a moderator:

#### lurflurf

Homework Helper
Line said:
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?
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.

#### Line

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

#### VietDao29

Homework Helper
HallsofIvy said:
(and, just to get back at you- 3(x-1)+ 1= 3x-3+1= 3x- 2 not 3x+2!) Yup. Thanks for pointing that out... Line said:
SO is it 3x-2 or 3x+2?
It's 3x - 2.
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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.
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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,

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