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find y' of sin((x^2)+(y^2))=4xy. I have already used the chain rule and product rule to get cos((x^2)+(y^2))(2x+2y*y')=4xy'+4y, I just need help isolating y'
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\[\cos(x ^{2}+y ^{2})(2x+2yy')=4xy'+4y\]
sin( (x^2)+(y^2) ) = 4xy ((x^2)+(y^2))' cos( (x^2)+(y^2) ) = 4y + 4xy' 2(x+yy') cos( (x^2)+(y^2) ) = 4y + 4xy' (x+yy') cos( (x^2)+(y^2) ) = 2y + 2xy' =A (x+yy')A = 2y + 2xy' distribute, and collect your like terms ...
You would distribute the A? the A is just a placeholder for cos( (x^2)+(y^2) ) right?
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