Find an equation of f

The slope at each point (x, y) and the graph of y = f (x) is given by x[x^(2)-1]^(1/3) and the graph passes through the point (3,1). Find the equation for f?

step by step please, I am lost.

This is a separable differential equation...

slope = dy/dx = x(x^2 - 1)^(1/3)

dy = x(x^2 - 1)^(1/3) dx

we need to integrate both sides....

let u=x^2 -1
then du/2 = x dx

plug that into the right side...
integrate 0.5u^(1/3) du

(3/8) u^(4/3) + C
plug "u" back in...
y = (3/8) (x^2-1)^(4/3) + C

To find "C" plug in the point...

1 = (3/8)(3^2-1)^(4/3) + C
C = -5.


So your answer is....

y = (3/8) (x^2-1)^(4/3) -5

This is an integration problem. To do it, you have to look at some function that will differentiate into y' = x[x^2-1]^1/3

Let u = x^2 - 1
du = 2x*dx
dx = du/2x

So integrate u^1/3 du/2

This gives u^(4/3)/[2*4/3] = (3/8)*u^4/3 + C

Which when you substitute for u gives y = (3/8) (x^2 -1)^(4/3) + C

Now we can solve for C
y = 1
x = 3
1 = (3/8) (8)^(4/3) + C
1 = (3/8) * 16 + C
1 = 6 + C
C = - 5

So the final answer is
y = (3/8) (x^2 - 1) ^ (4/3) - 5

The slope at x is f '(x)
therefore
f(x) = ∫x[x^(2)-1]^(1/3) dx = (*)

set
x^2 - 1 = u
2x dx = du
x dx = du/2

(*) = (1/2)∫u^(1/3) du = (1/2) (1/(1/3+1)) u^(1/(1/3+1)) + C
= (3/8)u^(3/4) + C = (3/8)(x^2 - 1)^(4/3) + C

f(3) = 1

f(3) = (3/8)(3^2 - 1)^(4/3) + C = 1

(3/8)8^(4/3) + C = 1
C= 1 - (3/8)8^(4/3) = 1 - (3/8)· 8· (8^1/3) = 1 - 3· 2 = -5

f(x) = (3/8)(x^2 - 1)^(4/3) - 5

f(x) = (3/8)[x^(2)-1]^(4/3) + C , plug (3,1) to find C