Question

How do you implicitly differentiate `-4=(2x+3y)^2-y^2x`?

Answer 1

`y'=(y^2-12y-8x)/(2(6x+9y-xy))`

Explanation:

We start from the given equation and differentiate both sides of the equation with respect to x

`-4=(2x+3y)^2-y^2x`

`d/dx(-4)=d/dx((2x+3y)^2-y^2x)`

`0=2(2x+3y)^(2-1)*d/dx(2x+3y)-d/dx(y^2x)`

`0=2(2x+3y)*(2+3y')-[y^2*1+x*2y*y'])`

Simplify at this point to obtain the `y'`

`0=(4x+6y)*(2+3y')-y^2-2xyy'`

`0=(8x+12y+12xy'+18yy')-y^2-2xyy'`

`0=8x+12y+12xy'+18yy'-y^2-2xyy'`

`(y^2-12y-8x)=(12x+18y-2xy)y'`

`y'=(y^2-12y-8x)/(12x+18y-2xy)`

`y'=(y^2-12y-8x)/(2(6x+9y-xy))`

God bless....I hope the explanation is useful.