If xy is a function of x^3, what is the slope of xy = ax^3 + b on this coordinate system?

Question

Y = mX + c

How come the slope of xy = ax^3 + b is not = a when the function is plotted in a graph of xy against x^3?

The original function was y = ax^2 + b/x. It was converted from a non-linear to linear function.

Thanks! I'm really confused

https://answers.yahoo.com/question/index?qid=20171107135543AAhNvyl

Answer 1

　

xy = f(x³) = ax³ + b



Since xy is a function of x³, not x, then using first principles we get:

(xy)' = lim[h→0] [f(x³+h) − f(x³)] / h

(xy)' = lim[h→0] [(a(x³+h)+b) − (ax³+b)] / h

(xy)' = lim[h→0] (ax³ + ah + b − ax³ − b) / h

(xy)' = lim[h→0] ah / h

(xy)' = a



Alternate method:



Find the slope between any points (x₁, y₁) and (x₂, y₂) on the x³-xy coordinate system:

Slope = Δxy/Δx³ = ((ax₁³+b)−(ax₂³+b)) / (x₁³−x₂³) = a(x₁³−x₂³)/(x₁³−x₂³) = a

Answer 2

Did you get a straight line for your plot of xy vs x^3? You should. Here's a parametric plot of t*(5t^2 + 3/t) vs t^3 as an example:

http://www.wolframalpha.com/input/?i=parametric+plot+(t%5E3,+t*(5t%5E2+%2B+3%2Ft)),+t%3D1..2



d(xy)/d(x^3) = [d(xy)/dx] / [d(x^3)/dx]

= {d[x(ax^2 + b/x)]/dx } / (3x^2)

= [d(ax^3 + b)/dx] / (3x^2)

= (3ax^2)/(3x^2)

d(xy)/d(x^3) = a .... yes, a is the constant slope of that graph



Try it with discrete points 1, x:



m = [xy(x) - 1y(1)] / (x^3 - 1^3) ..... (change in xy) / (change in x^3)

= [(ax^3 - b) - (a - b)] / (x^3 - 1)

= (ax^3 - a) / (x^3 - 1)

= a



Again, the slope is constant and equal to a.

Answer 3

The slope of xy is the derivative of x*(ax^2 + b/x) = derivative of ax^3 + b which is 3ax². It is thus not "a".