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what is dx/dy of 2x-5x^3y^2+4y=4?
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is the middle term 5x^3 and y^2
yes
so then you have to do the product rule of that middle section
so fg'+f'g
2x-[5x^3*dy/dx(2y)+(15x^2)(y^2)]+dy/dx(4y)=4
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then you have to isolate dy/dx
I was screwing up the 2nd part of the problem. I wasn't doing the y^2
are you good now?
I think so. Thanks a bunch. I should have done this 25 years ago when I knew this stuff.
yea just isolate the dy/dx, implicit differentiation isnt too bad. You should be fine now
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