How do you differentiate dy/dx with respect to y???

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MrChemKid
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As above .
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RDKGames
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(Original post by MrChemKid)
As above .
Nice question.

You use implicit differentiation from C4 which comes from the chain rule that you learn.

Let y=f(x)

Then \frac{dy}{dx}=f'(x)

\displaystyle \frac{d}{dy}(\frac{dy}{dx})= \frac{d}{dy} [f'(x)]

\displaystyle \frac{d}{dy}(\frac{dy}{dx})= \frac{dx}{dy} \cdot \frac{d}{dx}[f'(x)]

\displaystyle \frac{d}{dy}(\frac{dy}{dx})= f''(x)\frac{dx}{dy} = \frac{d^2y}{dx^2}\cdot \frac{dx}{dy}
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NotNotBatman
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(Original post by MrChemKid)
As above .
 \frac{d}{dy}(\frac{dy}{dx}) = \frac{d}{dx}(\frac{dy}{dx})\cdot  \frac{dx}{dy} = \frac{d^2y}{dx^2}\cdot \frac{dx}{dy}
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MrChemKid
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cheers
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3121
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I'm surprised you understood it with the formula, for me I had to practice it and understand the process before I could link it to the formula. To be honest even I do it now its more of an automated process rather than following a formula which is bad practice but its just a nasty thing to remember
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Donmclexy
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How do I differentiate dx/dy with respect to x
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RDKGames
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(Original post by Donmclexy)
How do I differentiate dx/dy with respect to x
Chain rule:

\dfrac{d}{dx}\left( \dfrac{dx}{dy} \right) = \dfrac{dy}{dx} \dfrac{d}{dx}\left( \dfrac{dx}{dy} \right)
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