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It's not a particularly challenging derivative, but I would like to know whether or not my approach is correct.

I assumed, prior to the process of differentiation that $y$ is a continuous function of $x$, and hence I applied the chain rule in the instances where $y$ occurred, and also therefore differentiated every term with respect to $x$

So, let's begin:

Term $1$: $\displaystyle\frac{d}{dx}\left(\frac{x}{1-x}\right)=(1-x)^{-2}$

Term $2$: At this point, I did the following: I let $u$ be a continuous function of $x$ such that $u=y^2$, and hence I applied the chain rule, which states that $\displaystyle \frac{du}{dx}=\frac{du}{dy}\times\frac{dy}{dx}$, hence $\displaystyle \frac{d}{dx}(y^2)=2y\frac{dy}{dx}$

Term $3$: $\displaystyle \frac{d}{dx}(3x^3)=9x^2$

Term $4$: Again, the chain rule is applied to attain $\displaystyle \frac{d}{dx}(5y)=5\frac{dy}{dx}$

And from here I solved for $\displaystyle \frac{dy}{dx}$, which is:

$\displaystyle (1-x)^{-2}+9x^2=5\frac{dy}{dx}-2y\frac{dy}{dx}$

$\displaystyle\frac{dy}{dx}=\frac{9x^2}{(5-2y)(1-x)^2}$

So here are the questions; firstly, when differentiating implicit functions, will you always differentiate with respect to only one variable? Secondly, for the chain rule, is it a good idea to substitute, for example; $u=y^2$ or $n=5y$ and then calculate $\displaystyle \frac{du}{dx}$ or $\displaystyle \frac{dn}{dx}$, each time I come across a variable that is not $x$ (in this example)? And lastly, will this approach work for the differentiation of most, if not every, implicit function?

Any responses are appreciated.

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  • $\begingroup$ Are we sure that $\dfrac{d}{dx} \dfrac{x}{1-x}=(1-x)^{-2}$ I wasn't able to check with pen and paper but it looks to me like it is $0$ if we apply the quotient rule. $\endgroup$ – Deniz Tuna Yalçın Sep 1 '17 at 7:39
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    $\begingroup$ Yes you were right OK $\endgroup$ – Deniz Tuna Yalçın Sep 1 '17 at 7:40
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    $\begingroup$ For your second question about substituting I think it is safer to do it that way but I feel like it is easier to use $y'$ instead of the d notation and skip the intermeadiate steps like (say) $x^2+y^2=1 \rightarrow 2x+2yy'=0$ (easier than substituting although riskier. $\endgroup$ – Deniz Tuna Yalçın Sep 1 '17 at 7:43
  • $\begingroup$ Makes sense, thanks for the advice! @DenizTunaYalçın $\endgroup$ – joshuaheckroodt Sep 1 '17 at 7:44
  • $\begingroup$ You're welcome:)) $\endgroup$ – Deniz Tuna Yalçın Sep 1 '17 at 7:45
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Your method is correct, but they are some calculus mistakes at the end :

enter image description here

The same process but on another form :

$$\frac{x}{1-x}-y^2+3x^3=5y \tag 1$$ $$\left(\frac{1}{1-x}+\frac{x}{(1-x)^2}\right)dx -2y\:dy+9x^2dx=5dy \tag 2$$ $\frac{1}{1-x}+\frac{x}{(1-x)^2} = \frac{1}{(1-x)^2}$ $$ (2y+5)dy=\left(\frac{1}{(1-x)^2}+9x^2\right)dx$$ $$ (2y+5)dy=\frac{1+9x^2(1-x)^2}{(1-x^2)}dx$$ $$y'=\frac{dy}{dx}=\frac{1+9x^2(1-x)^2}{(1-x^2)(2y+5)}$$

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  • $\begingroup$ Could you potentially add some word-based explanation to the steps of your process? $\endgroup$ – joshuaheckroodt Sep 1 '17 at 18:00
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    $\begingroup$ If you known the rules for differentiation, all is obvious. The second line is the Total Differential of the first line. Do you know what is the Total Differential of a function ? If not : en.wikipedia.org/wiki/… $\endgroup$ – JJacquelin Sep 1 '17 at 18:10
  • $\begingroup$ I see, thank you for the help :) $\endgroup$ – joshuaheckroodt Sep 1 '17 at 18:17

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