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need to differentiation twice here, y''.
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?
this is question:
sorry idk :/
alright, thanks for looking... anyone else can help?
\[2\sqrt y = x-y\]
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Ok Lets Differentiate first time \[\frac{1}{\sqrt y} \frac{dy}{dx} = 1 - \frac{dy}{dx}\]
then
Now Differentiate again \[-\frac{1}{2 y^3/2} \frac{dy}{dx} + \frac{d^2y}{dx^2} \frac{1}{\sqrt y} = - \frac{d^2y}{dx^2}\]
There Is your Double Differentiate
well it is \[\Large {-\frac {1}{2y^{ \frac {3}{2} }}\frac{ dy}{dx} } \]
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I made a typo there
is that answer = to dy/dx? or is that the final answer as it is..?
No it is the final answer if you want you can have \[\frac{d^2y}{dx^2}\]on one side and have your answer or just let it remain like this
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