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

• Line
In summary: Oops! In summary, the slope of a tangent can have an exponent, but why? 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. The linearity is in another variable. So is it 3x-2 or 3x+2? (and, just to get back at you- 3(x-1)+ 1= 3x-3+1= 3x- 2 not 3x+2!)
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?

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.

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.

OK how did you get 3X-2?

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:
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:
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.

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

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.
------------------------
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,

What are exponents on slopes?

Exponents on slopes are a way of representing a mathematical relationship between two variables, where one variable is raised to a certain power. It is often used in algebra and calculus to express growth or decay rates.

How do you calculate exponents on slopes?

To calculate exponents on slopes, you need to determine the base value and the exponent. Then, you can use the formula y = mx^b, where y is the dependent variable, x is the independent variable, m is the slope, and b is the exponent. Plug in the values and solve for y to find the corresponding points on the slope.

What is the significance of exponents on slopes?

Exponents on slopes are important in understanding the rate of change or growth of a variable. They can also be used to make predictions about future values based on the current trend.

How is the slope affected by the exponent?

The slope is affected by the exponent in a few ways. First, a larger exponent will result in a steeper slope, indicating a faster rate of change. Second, a negative exponent will result in a downward slope, indicating a decrease in the variable. Lastly, a slope of zero (when the exponent is 0) indicates a constant value.

Can exponents on slopes be negative?

Yes, exponents on slopes can be negative. This means that the variable is decreasing at a certain rate. However, it is important to note that the base value must be positive in order for the slope to be negative. A negative base value with a negative exponent would result in a positive slope.

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