How do you find dy/dx implicitly of #ye^(3x) - 4^x = 2y^2#?

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Answer 1 · 2015-05-06T16:47:22

#ye^(3x) - 4^x = 2y^2#

To differentiate we'll need the product rule for the first term.

#d/dx(ye^(3x) - 4^x)= d/dx(2y^2)#

#dy/dx e^(3x) + y e^(3x) 3 = 4y dy/dx#. So

# e^(3x) dy/dx + 3y e^(3x) = 4y dy/dx#.

Solve for #dy/dx#:

# dy/dx = (3y e^(3x)) /(4y - e^(3x))#.