# simplify implicit differentiation equation

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 :)

• there is a typo in your equation – Nosrati Sep 16 '17 at 18:17
• Is $x+x$ a typo? You have also split the fraction on the LHS incorrectly. – George Coote Sep 16 '17 at 18:18
• yes thanks its been corrected. – user2355058 Sep 16 '17 at 18:19

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 $$y'=\frac{7xye^{7x-y}-x-y}{1+xye^{7x-y}}$$
• it is corrected, instead of $2x$ it is $x+y$ now – Dr. Sonnhard Graubner Sep 16 '17 at 18:27
• @user2355058 He factored $y'$, ie. $$y' + y'xye^{7x-y} \equiv y'(1+xye^{7x-y})$$ at which point you can divide – George Coote Sep 16 '17 at 18:35