Proving an equation related to order of derivatives

[madah12]

Homework Statement

Prove that for all y(x)=ax^2+bx+c where a is a constant !=0 and x is a real number that \frac{y'(x2)^2- y'(x1)^2}{(x2-x1)} = 2y''(x)

Homework Equations

I don't know what to put here in mathematics but here...
y(x)=ax^2+bx+c
y'(x)=2ax+b
y''(x)=2a

The Attempt at a Solution

I made a lot of attempts and tried many thing but here is the one that looks like it works.
y''(x)=dy'(x)/dx
y''(x) * dy/dx = y'(x)dy'/dx
now here I don't know if its ok to take the dx es of the fraction
y''(x) dy= y'(x)dy'
integral from x1 to x2 (y''(x) dy)= integral from x1 to x2 (y'(x)dy')
and since y''(x)=2a so its a constant
y''(x)* (x2-x1)= 1/2 * (y'(x2)^2-y'(x1)^2)
2y''(x)= (y'(x2)^2-y'(x1)^2)/ (x2-x1)

 

[Mark44]

Homework Statement

Prove that for all y(x)=ax^2+bx+c where a is a constant !=0 and x is a real number that \frac{y'(x2)^2- y'(x1)^2}{(x2-x1)} = 2y''(x)

This is ambiguous. What does y'(x₂)² mean? Is it the derivative evaluated at (x₂)² or is it the square of the derivative evaluated at x₂?

Homework Equations

I don't know what to put here in mathematics but here...
y(x)=ax^2+bx+c
y'(x)=2ax+b
y''(x)=2a

The Attempt at a Solution

I made a lot of attempts and tried many thing but here is the one that looks like it works.
y''(x)=dy'(x)/dx

You already know that y''(x) = 2a, so I don't see that this or the next line are of any help.

y''(x) * dy/dx = y'(x)dy'/dx
now here I don't know if its ok to take the dx es of the fraction
y''(x) dy= y'(x)dy'
integral from x1 to x2 (y''(x) dy)= integral from x1 to x2 (y'(x)dy')
and since y''(x)=2a so its a constant
y''(x)* (x2-x1)= 1/2 * (y'(x2)^2-y'(x1)^2)
2y''(x)= (y'(x2)^2-y'(x1)^2)/ (x2-x1)

I'm not sure that what you have above is correct. In any case, it's a lot more convoluted than it needs to be.

 

[madah12]

This is ambiguous. What does y'(x₂)² mean? Is it the derivative evaluated at (x₂)² or is it the square of the derivative [\QUOTE]
It means the square of the derivative evaluated at x2

 

[Mark44]

Then to be clear, you should write (y'(x₂)².

Evaluate the left side:
\frac{(y'(x_2))^2 - (y'(x_1))^2}{x_2 - x_1}

What does it take for the simplified left side to equal 4a (= 2y''(x))? Aside from taking the first and second derivatives of the given function, this problem seems to me to be an algebra exercise.

 

[madah12]

Then to be clear, you should write (y'(x₂)².

Evaluate the left side:
\frac{(y'(x_2))^2 - (y'(x_1))^2}{x_2 - x_1}

What does it take for the simplified left side to equal 4a (= 2y''(x))? Aside from taking the first and second derivatives of the given function, this problem seems to me to be an algebra exercise.

But is the solution I already wrote incorrect?

 

[Mark44]

It took me awhile to figure out what you were doing, but it seems to be OK.

 

[madah12]

It took me awhile to figure out what you were doing, but it seems to be OK.

I don't understand how you solved it with algebra. since my solution is ok it would be ok for you to post the algebriac solution right? it won't be cheating for me since I already made my own solution right?

 

[Mark44]

Right now, I like your approach better than mine. Let me think about it a bit.

 

[vela]

You're integrating the lefthand side with respect to y, so you should get 2a(y₂-y₁), not 2a(x₂-x₁).

The original statement is false. If you have y=x² for instance, you get

\frac{[y'(x_2)]^2-[y'(x_1)]^2}{x_2-x_1} = 4(x_2+x_1)

You're not going to be able to prove it.

 

[madah12]

You're integrating the lefthand side with respect to y, so you should get 2a(y₂-y₁), not 2a(x₂-x₁).

The original statement is false. If you have y=x² for instance, you get

\frac{[y'(x_2)]^2-[y'(x_1)]^2}{x_2-x_1} = 4(x_2+x_1)

You're not going to be able to prove it.

oh sorry the denominator was y(x2)-y(x1) I was translating it from a different language with different variables so...
Edit:
also I think this is proven
in the dimensional motion
(V(t))^2=V(0))^2 + 2a (x2-x1) I just noticed lol think about it
v= y'
and a=y'' and x=y then it would all make sense lol

 

[Mark44]

You're integrating the lefthand side with respect to y, so you should get 2a(y₂-y₁), not 2a(x₂-x₁).

The original statement is false. If you have y=x² for instance, you get

\frac{[y'(x_2)]^2-[y'(x_1)]^2}{x_2-x_1} = 4(x_2+x_1)

You're not going to be able to prove it.

Good catch, vela!

 

[madah12]

Good catch, vela!

I mistyped the question sorry but I thinks its fixed now