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now i guess i just have to apply the product rule in reverse ,

im not sure

Hmmm oh yeah, there is that \(x\) in there...

i dunno

What inspired the partial derivative, was it just a guess?

@wio im not sure, it worked on the previous question

Ah, I see. I was a big confused as to the fact that we have a multivariable function

Actually, the part titled "Formal Statement" on that page looks useful here...

What would \(b(\theta)\) ad \(\theta\) be?

\(b(\theta) = x \) and \(\theta = x\)?

@UnkleRhaukus did you look at the formal statement?

yeah , but im not sure how to use that in my problem

Did you try writing it out to see what happens?

@AccessDenied actually it seems when you write it out, the added parts go to 0

Yeah, I see that now. Interesting. :)

@UnkleRhaukus @AccessDenied could you clear this up?

How can a single variable function have a partial derivative?

imagine if \(f(x) = e^x\) the \(f(u)=e^u\) right?

um

im not sure

If you do this: \[ \large
\frac{dy}{dx} = \int^x_0 f(u)\cos k(x-u) du
\]

then try the thing that @AccessDenied mentioned.. I wonder

\[\large
\frac{d^2y}{dx^2} = -k\int^x_0 f(u)\sin k(x-u) du +f(x)
\]
Oh pellet, it works!

@UnkleRhaukus All you gotta do is treat \(f(u)\) as a constant and it'll work.

i dont understand