How do you differentiate #y^2cos(x) = x/(x-y)#?

1 Answer
May 18, 2016

See below.

Explanation:

In terms of differentiating the above equation, #dy/dx# is to be found.

If you differentiate both sides and fiddle a little bit, it's possible to extract #dy/dx#.

The left side will need to be differentiated with the product rule and the left with the quotient rule.

Furthermore, taking derivatives of functions of y occur as follows:

#(df(y))/dx=(df(y))/dy dy/dx#, by the chain rule.
This can be simply remembered as just differentiating the function with respect to #y# and just sticking a #dy/dx# on the end.

Let's differentiate the left side first:
#d/dx(y^2cosy)=d/dx(y^2)cosy+d/dx(cosy)y^2#
#d/dx(y^2cosy)=2ycosydy/dx-y^2sinydy/dx#

The right side:
#d/dx(x/(x-y))=(d/dx(x)(x-y)-d/dx(x-y)x)/(x-y)^2#
#d/dx(x/(x-y))=(x-y-x(1-dy/dx))/(x-y)^2#
#d/dx(x/(x-y))=(y+xdy/dx)/(x-y)^2#

Therefore:
#y^2cosy=x/(x-y)#
Becomes:
#2ycosydy/dx-y^2sinydy/dx=(y+xdy/dx)/(x-y)^2#
We need to group all of the #dy/dx# parts on one side in order to solve for it:
#2ycosydy/dx-y^2sinydy/dx-(xdy/dx)/(x-y)^2=y/(x-y)^2#
Factorise by #dy/dx#:
#dy/dx(2ycosy-y^2siny-x/(x-y)^2)=y/(x-y)^2#
Therefore (and this is a little messy):
#dy/dx=y/((x-y)^2(2ycosy-y^2siny-x/(x-y)^2))#
Tidy it up a little bit:
#dy/dx=y/((x-y)^2(2ycosy-y^2siny)-x)#