can someone solve this problem for me ?
find dy/dx (x,y)=(3,3) by implicit differentiation given: x^3+y^3=6xy

Question

can someone solve this problem for me ?
find dy/dx (x,y)=(3,3) by implicit differentiation given: x^3+y^3=6xy

sidsiddhartha · Answer

just differentiate both sides using chain rule

perl · Answer

3x^2  + 3y^2 * y ' = 6* ( 1*y + x*y' )

anonymous · Answer

Im having trouble understanding implicit differentiation. I make it all the way to the point where I have to separate the dy/dx... I thought seeing someone else's work would help me see where I was going wrong... ;/

sidsiddhartha · Answer

the function is
$$x^3+y^3=6xy\differentiate~both~sides\3x^2+3y^2*\frac{ dy }{ dx }=6x \frac{ dy }{ dx }+6y$$=

sidsiddhartha · Answer

ok? do u understand what i did

anonymous · Answer

The steps I do.
i'm confused on why you kept the dy/dx on the 3y^2 and 6x... 
btw, thanks for helping me

sidsiddhartha · Answer

ok when you are differentiating y terms with respect to x, u have to include a dy/dx with it

sidsiddhartha · Answer

take a simple example
$$f(x)=siny\then\f'(x)=cosy*\frac{ dy }{ dx}$$

sidsiddhartha · Answer

like this ok?

anonymous · Answer

Yeah! Thanks!

sidsiddhartha · Answer

can u do the rest? its just algebra ok!!

sidsiddhartha · Answer

take terms with dy/dx in any side then take common

anonymous · Answer

yes, i worked out the right answer :)

sidsiddhartha · Answer

well done :)

perl · Answer

@sidsiddhartha
sid theres a mistake 
you wrote f(x) = sin y ,  
it should be f(y) = sin y

sidsiddhartha · Answer

yup my bad it will be f(y)=siny as siny is a function of y 
thanks @perl

perl · Answer

given 
f(y(x))

(f(y(x)) ' = f ' (y(x)) * y' (x)  

or more simply 

[f(y)] ' =  f ' (y) * dy/dx

perl · Answer

no problem :D