implicit differentiation of x^2 + 3xy + y^3 = 10

[charnitr]

x[sup:2o0vqx78]2[/sup:2o0vqx78]+3xy+y[sup:2o0vqx78]3[/sup:2o0vqx78]=10

i know that you are supposed to take the derivative (dy/dx) and use the product rule were needed
so you would do..
2x+ [(3x*y')+(3*y)]+3y[sup:2o0vqx78]2[/sup:2o0vqx78]=0

im not quite sure where to go from there

answer: -2x+3y/3(x+y[sup:2o0vqx78]2[/sup:2o0vqx78])

 

[mmm4444bot]

Re: implicit differentiation

... 2x + [(3x*y') + (3*y)] + 3y[sup:2xc33119]2[/sup:2xc33119] = 0 ? The derivative of y[sup:2xc33119]3[/sup:2xc33119] with respect to x is not 3y[sup:2xc33119]2[/sup:2xc33119].

The chain rule tells us that this derivative is 3y[sup:2xc33119]2[/sup:2xc33119]y'.

... answer: -[2x + 3y]/[3(x + y[sup:2xc33119]2[/sup:2xc33119])] ? Use grouping symbols to clearly show numerators and denominators.

The answer is not correct with +3y in the numerator, unless you type grouping symbols to show that the negative ones are factored out.

After correcting the derivative of y[sup:2xc33119]3[/sup:2xc33119], solve the equation for y'.

2x + 3xy' + 3y + 3y[sup:2xc33119]2[/sup:2xc33119]y' = 0

Cheers,

~ Mark

 

[qpmathelp]

diff. the given expression w.r.t.x
2x + 3[x (dy/dx) + y.1] + 3y² (dy/dx) = 0

(dy/dx) [3x + 3y²] = -2x -3y

solve for dy/dx