Why can the tangent of a curve have exponents in Calculus?

[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]

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]

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]

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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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,

 

[HallsofIvy]

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!)

 

[lurflurf]

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]

(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...

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

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,