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I have a few implicit diffentiatial questions that I wanted to check. general question... how do you know that $y$ is a function of $x$? I assume the whole reason why these are called implicit differentiation questions is because $y$ is defined implicitly and it's hard or impossible to define $y$ as a function of $x$. Is that right?

1. $$ \cos{xy} = 1+ \sin{y} $$ $$-\sin{xy} \cdot (\frac{dy}{dx} + y) = \cos{y} \cdot \frac{dy}{dx}$$ $$-\sin{xy} \cdot \frac{dy}{dx} - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx}$$ $$ - y(\sin{xy}) = \cos{y} \cdot \frac{dy}{dx} + \sin{xy} \cdot \frac{dy}{dx}$$ $$ - y(\sin{xy}) = \frac{dy}{dx} ( \sin{xy} + \cos{y})$$ $$ \frac{- y(\sin{xy})}{( \sin{xy} + \cos{y})} = \frac{dy}{dx}$$

  1. $$x - y = x \cdot e^y$$ $$1 - \frac{dy}{dx} = x \cdot e^y \cdot \frac{dy}{dx} + e^y$$ $$ 1 - e^y = x \cdot e^y \frac{dy}{dx} + \frac{dy}{dx}$$ $$1 - e^y = \frac{dy}{dx}(x \cdot e^y + 1)$$ $$\frac{1-e^y}{x \cdot e^y + 1} = \frac{dy}{dx}$$

  2. $$y \cdot \cos{x} = x^2 + y^2$$ $$y(-\sin{x}) \cdot \cos{x} \cdot \frac{dy}{dx} = 2x + 2y \frac{dy}{dx}$$ $$ 2y \frac{dy}{dx} - \cos{x} \frac{dy}{dx} = y - \sin{x} - 2x$$ $$\frac{dy}{dx} (2y-cosx) = y - sinx - 2x$$ $$\frac{dy}{dx} = \frac{y - sinx - 2x}{2y - cosx}$$

Thank you.

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    $\begingroup$ There is an error on your second line where you should have $x\frac{dy}{dx}+y$ rather than $\frac{dy}{dx}+y$. $\endgroup$ – John Wayland Bales Feb 6 '18 at 0:35
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1) As John pointed there is a mistake...

2) The second example seems correct to me

3) For the third differenciation... $$y \cdot \cos{x} = x^2 + y^2$$ $$y'\cos(x)-y\sin(x)=2x+2yy'$$ $$y'\cos(x)-2yy'=y\sin(x)+2x$$ $$y'(\cos(x)-2y)=y\sin(x)+2x$$ $$\frac {dy}{dx}=\frac {y\sin(x)+2x}{\cos(x)-2y} $$ ...

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    $\begingroup$ ohhhh I confused myself and thought a negative sign was a minus sign next to sinx. My handwriting sucks. $\endgroup$ – Jwan622 Feb 6 '18 at 1:11
  • $\begingroup$ @Jwan622 oh it heppens to me too all the time lol $\endgroup$ – Isham Feb 6 '18 at 1:34
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Note that in the case of implicit differentiation it is not necessarily the case that one variable is a function of the other in the strict sense, but the two functions are related. One can symbolize this as $x$ and $y$ being parametric functions of a variable $t$. An alternate approach to implicit differentiation is based on this idea.

Taking your first example:

\begin{eqnarray} \cos{xy} &=& 1+ \sin{y}\\ \frac{d}{dt}\left(\cos{xy} \right)&=&\frac{d}{dt}\left(1+ \sin{y}\right)\\ -\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)\sin xy&=&(\cos y)\frac{dy}{dt}\\ -(x\,dy+y\,dx)\sin xy&=&\cos y\,dy\\ -y\sin xy\,dx&=&(x\sin xy+\cos y)\,dy\\ \frac{dy}{dx}&=&-\frac{y\sin xy}{x\sin xy+\cos y} \end{eqnarray}

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