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I am a Calculus 2 student. I am doing implicit differentiation and I want to know the fastest way to simplify this and find y'. My online algebra calculator fails to ever solve problems the easy way. Hoping a math pro on here could show me "the easy way".

$$ \frac{x+y+y'}{xy}=e^{7x-y}(7-y') $$

$$ = $$

$$ \frac{1}{x}+\frac{y'}{y}=7e^{7x-y}-y'e^{7x-y} $$

$$ y'=? $$

Adding and subtracting messy fractions is never fun. Help friends :)

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  • $\begingroup$ there is a typo in your equation $\endgroup$ – Nosrati Sep 16 '17 at 18:17
  • $\begingroup$ Is $x+x$ a typo? You have also split the fraction on the LHS incorrectly. $\endgroup$ – George Coote Sep 16 '17 at 18:18
  • $\begingroup$ yes thanks its been corrected. $\endgroup$ – user2355058 Sep 16 '17 at 18:19
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multiplying by $$xy$$ we get $$x+y+y'=xye^{7x-y}(7-y')$$ multiplying out: $$x+y'+y=7xye^{7x-y}-y'xye^{7x-y}$$ $$y'+y'xye^{7x-y}=7xye^{7x-y}-x-y$$ $$y'(1+xye^{7x-y})=7xye^{7x-y}-x-y$$ therefore $$y'=\frac{7xye^{7x-y}-x-y}{1+xye^{7x-y}}$$

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    $\begingroup$ The OP has modified their question, hence you should change your answer a bit. $\endgroup$ – George Coote Sep 16 '17 at 18:24
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    $\begingroup$ it is corrected, instead of $2x$ it is $x+y$ now $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 18:27
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    $\begingroup$ and my Point is going down now? $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 18:31
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    $\begingroup$ @Dr.SonnhardGraubner sorry, I leant on the (un)-upvote button by mistake $\endgroup$ – George Coote Sep 16 '17 at 18:32
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    $\begingroup$ @user2355058 He factored $y'$, ie. $$y' + y'xye^{7x-y} \equiv y'(1+xye^{7x-y})$$ at which point you can divide $\endgroup$ – George Coote Sep 16 '17 at 18:35

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