Calculus Proof, please help!?

Question

i am studying for a test and cant figure out this question:
if y=(2x-1)^4, prove:
d/dx (y* dy/dx) = y * (d^2*y) / dx^2 + (dy/ddx)^2

hope that makes sense
thanks in advance

Answer 1

Well, y'=8(2x-1)^3 - we'll need this.

y'' = 48(2x-1)^2 - gonna need this too.

LHS: (yy')' = d/dx (8(2x-1)^7) = 112(2x-1)6
RHS: yy''+y'^2
= 48(2x-1)^6+64(2x-1)^6 = 112(2x-1)^6

So, yes, they are the same.

' used for d/dx for convenience

Answer 2

Basically this is the an example of the general rule.

When given 2 functions, say f(x) and g(x), than
d/dx(f*g)=(d/dx f)*g + f * (d/dx g)

were f=y and g=dy/dx

in your case you are asked to demonstrate it explicitly:

d/dx y = 4(2x-1)^3*2=8(2x-1)^3
d^2/dx^2 y = 24(2x-1)^2*2=48(2x-1)^2

Inserting these results into the LEFT hand side of the requested identity yields:
d/dx ((2x-1)^4*8(2x-1)^3)
=d/dx (8(2x-1)^7)
=56(2x-1)^6*2
=112(2x-1)^6

Inserting the derivative results into the RIGHT hand side yields:
(2x-1)^4*48(2x-1)^2+(8*(2x-1)^3)^2
=48(2x-1)^6+64(2x-1)^6
=112(2x-1)^6

Answer 3

So we're looking to show that:

d/dx ( y * dy/dx) = y * d²y/dx² + (dy/dx)²

I've chosen to find each of the terms first, then combine them.

y = ( 2x - 1 )^4
dy/dx = 4*( 2x - 1 )^3 * 2 = 8*( 2x - 1 )^3
d²y/dx² = 8*3*( 2x - 1 )^2 * 2 = 48*( 2x - 1 )^2

y * dy/dx = ( 2x - 1 )^4 * 8*( 2x - 1 )^3 = 8*( 2x - 1 )^7
y * d²y/d²x = ( 2x - 1 )^4 * 48*( 2x - 1 )^2 = 48*( 2x - 1 )^6

d/dx ( y * dy/dx ) = d/dx ( 8*( 2x - 1 )^7 ) = 8*7*( 2x - 1 )^6 * 2 = 112*( 2x - 1 )^6

(dy/dx)² = ( 8*( 2x - 1 )^3 )^2 = 64*( 2x - 1 )^6

So, combining them:

d/dx ( y * dy/dx) = y * d²y/dx² + (dy/dx)²
112*( 2x - 1 )^6 = 48*( 2x - 1 )^6 + 64*( 2x - 1 )^6
112*( 2x - 1 )^6 = ( 48 + 64 )*( 2x - 1 )^6
112*( 2x - 1 )^6 = 112*( 2x - 1 )^6

It works!

Answer 4

y = ( 2x - 1 )^4
dy/dx = 4*( 2x - 1 )^3 * 2 = 8*( 2x - 1 )^3
d²y/dx² = 8*3*( 2x - 1 )^2 * 2 = 48*( 2x - 1 )^2

y * dy/dx = ( 2x - 1 )^4 * 8*( 2x - 1 )^3 = 8*( 2x - 1 )^7
y * d²y/d²x = ( 2x - 1 )^4 * 48*( 2x - 1 )^2 = 48*( 2x - 1 )^6

d/dx ( y * dy/dx ) = d/dx ( 8*( 2x - 1 )^7 ) = 8*7*( 2x - 1 )^6 * 2 = 112*( 2x - 1 )^6

(dy/dx)² = ( 8*( 2x - 1 )^3 )^2 = 64*( 2x - 1 )^6

d/dx ( y * dy/dx) = y * d²y/dx² + (dy/dx)²
112*( 2x - 1 )^6 = 48*( 2x - 1 )^6 + 64*( 2x - 1 )^6
112*( 2x - 1 )^6 = ( 48 + 64 )*( 2x - 1 )^6
112*( 2x - 1 )^6 = 112*( 2x - 1 )^6

Answer 5

d/dx(y * dy/dx) = y * d/dx(dy/dx) + dy/dx * dy/dx =
= y * (d^2 y)/dx^2 + (dy/dx)^2