How do you implicitly differentiate # y^2= x^3 + 3x^2 #?

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Answer 1 · 2015-11-20T15:55:07

It depends on what you are using as the independent variable. For #t#, we get #2y dy/dt=3x^2 dx/dt+6x dx/dt#.

# y^2= x^3 + 3x^2 #

#d/dt( y^2) = d/dt(x^3 + 3x^2)#

#2y dy/dt=3x^2 dx/dt+6x dx/dt#.

Solve for the derivative you're looking for.

If the independent variable is taken to be #x#, then we get:

# y^2= x^3 + 3x^2 #

#d/dx( y^2) = d/dx(x^3 + 3x^2)#

#2y dy/dx =3x^2+6x#

#dy/dx = (3x^2+6x)/(2y)#