find y' if y=ln(7x^2+3y^2)

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okay so i've asked this question before and i really appreciate the help you guys gave me. i want to see if what i've done so far is correct.

1st step: (7x^2+3y^2)'/ 7x^2+3y^2

2nd step: 14x+6yy'/ 7x^2+3y^2

3rd step: 14x/7x^2+3y^2 + 6yy'/7x^2+3y^2

4th step: y'- 6yy'/7x^2+3y^2= 14x/7x^2+3y^2

5th step: y'(1-6y/7x^2+3y^2)= 14x/7x^2+3y^2

6th step: y'((7x^2+3y^2)-6y)=14x

final step? y'= 14x/7x^2+3y^2-6y this is what i ended up with as my answer

also forgive me for my formatting.

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2 Answers

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Looks good! here is another way that slightly simplifies the arithmetic

$$ y = \ln(7x^2+3y^2) \implies e^y = 7x^2+3y^2 $$

Differentiating implicitly and isolating $y'$ gives you the same answer

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This is solution: enter image description here

if you will need more explain say.

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