∫y'(1/x^2)-∫2y/x^3 = y/x^2

You're right! My bad, I left the dx's out.

I'm not quite sure what you're meaning. The derivate of y/x^2 is -2y/x^3. But you probably meant something else. Give me another bone. :) English isn't my mother language so my terminolgy might be sometimes little off.
 
mikuumi said:
\(\displaystyle \displaystyle \int\, \left(y'\, \dfrac{1}{x^2}\right)\, dx\, -\, \int\, \left(\dfrac{2y}{x^3}\right)\, dx\, =\, \dfrac{y}{x^2}\)

I'm not quite sure what you're meaning. The derivative of y/x^2 is -2y/x^3.
Click to expand...
Shouldn't there be a "dy/dx" somewhere in your derivative?

mikuumi said:
But you probably meant something else.
Click to expand...
In light of what is discussed here, what happens if you differentiate the whole original equation? ;)
 
Oh boy! Here's step-by-step from my head so you guys can pinpoint where I miss the mark.

[FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]′ [/FONT][FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Math]x^2[/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]− [/FONT][FONT=MathJax_Size2]∫ [/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x^[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]^2

Integral and derivate cancel eac others

[FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Math]x^2[/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main]− [/FONT][FONT=MathJax_Size2]∫ [/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x^[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]^2

Then I'll integrate what is been subtracted[/FONT]

[FONT=MathJax_Main] [/FONT][FONT=MathJax_Size2]∫ [/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x^[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main] = -y/x^2

Then I have a simple equation which doesn't hold

[FONT=MathJax_Main][/FONT][FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Math]x^2[/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Math] - (-)y/x^2[/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Size2][/FONT][FONT=MathJax_Size3][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]^2

Thanks for beeing helpfull and patient.:D

p.s equation didn't work for me in the "equation editor form" which is a lot clealer nicer to look at.[/FONT]
[/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT]


[/FONT]
 
Last edited:
mikuumi said:
Oh boy! Here's step-by-step from my head so you guys can pinpoint where I miss the mark.

[FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]′ [/FONT][FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Math]x^2[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]− [/FONT][FONT=MathJax_Size2]∫ [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x^[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]^2

Integral and derivate cancel eac others[/FONT]

[FONT=MathJax_Main][FONT=MathJax_Main]1/[/FONT][FONT=MathJax_Math]x^2[/FONT][FONT=MathJax_Main]− [/FONT][FONT=MathJax_Size2]∫ [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x^[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]^2[/FONT][/FONT]
Click to expand...
How do you figure that they "cancel out"? For instance,

. . .\(\displaystyle \mbox{Let }\, f(x)\, =\, y\, =\, 3x,\, \mbox{ so }\, \dfrac{df}{dx}\, =\, y'\, =\, 3.\)

. . .\(\displaystyle \mbox{Then:}\)

. . . . .\(\displaystyle \displaystyle \int\, \left(y'\, \dfrac{1}{x^2}\right)\, dx\, =\, \int\, \left(\dfrac{3}{x^2}\right)\, dx\, \)

. . . . . . . .\(\displaystyle =\, -\dfrac{3}{x}\, +\, C\, \neq\, \dfrac{1}{x^2}\)

What is specific to whatever they told you about "y" which, in your case, somehow allows this "cancellation" thing to be done?

Please be complete. Thank you! ;)
 
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