Derivatives question

[Apricity]

Hi all!

Please help me with this question: "Given that y=x^2 - x, prove that f'(x) - f"(x) = (2y-x)/x

This is what I did:

f'(x)= 2x -1
f"(x)= 2

Then,
f'(x) - f"(x) = 2x -1 -2
= 2x -3

How do I get the (2y-x)/x ? Thanks very much for your help

 

[Deleted member 4993]

Hi all!

Please help me with this question: "Given that y=x^2 - x, prove that f'(x) - f"(x) = (2y-x)/x

This is what I did:

f'(x)= 2x -1
f"(x)= 2

Then,
f'(x) - f"(x) = 2x -1 -2
= 2x -3

How do I get the (2y-x)/x ? Thanks very much for your help

(2y-x)/x = [2x^2 - 2x -x]/x = [2x^2 - 3x]/x = x*(2x - 3)/x = 2x - 3

 

[Ishuda]

Or go the other way:
\(\displaystyle 2x-3=\frac{(2x-3) x}{x} = \frac{2x^2-3x}{x}= \frac{2x^2-2x-x}{x}= \frac{2(x^2-x)-x}{x}= \frac{2y-x}{x}\)

 

[Apricity]

Thank you very much to both of you