Break it up into its component parts and then put it all back together once differentiated.
#color(blue)("Consider "x^2")#
Using the generic form #y= x^n->(d(x^n))/dy = nx^(n-1)#
#color(brown)((d(x^2))/dy = 2x)#
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#color(blue)("Consider "2y^2)# This is an 'implicit' differentiation
We use the chain rule to 'force' what we have into what the target is.
Example: suppose we wish to end up with #a/c# but we have #a/b and b/c# Then if we do this: #a/bxxb/c = a/c# as required
Given that we do not know what the value of #dy/dx# is at the current time.
Differentiate the #y# term with respect to #y#
#(d(2y^2))/dy ->(2xx2)y^(2-1)=4y......Equation(1)#
But we need it to end up as the differential with respect to #x#
#dy/cancel(dy)xxcancel(dy)/dx -> dy/dx# so we change the way we handle #Equation(1)# as follows:
#(d(2y^2))/dy xxdy/dx = 4yxxdy/dx#
So #color(brown)((d(2y^2))/dx=4ydy/dx)#
where as yet we do not know what #dy/dx# is
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#color(blue)("Consider "2xy)# This is an 'implicit' differentiation but also uses the product rule:
Given that #y=uv# then #dy/dx = u(d(v))/dx+(d(u))/dxv#
You may have been shown the generic case in this format
#D(f(x)g(x))=f(x)g'(x)+f'(x)g(x)#
The thing is; #(d(v))/dx # is an implicit differentiation as we are differentiating #y# with respect to #y#
#dy/cancel(dy)xxcancel(dy)/dx=dy/dx#
Note that #(d(y))/dx = 1dy/dx = dy/dx#
So we have:
#(d(2xy))/dx = u(d(v))/dx+(d(u))/dxv color(white)("d") -> color(white)("d") 2[xdy/dx+y]#
#color(brown)((d(2xy))/dx=2xdy/dx+2y)#
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#color(blue)("Putting it all together")#
#(d(x^2+2xy+2y^2=1))/dx color(white)("d") -> color(white)("d") 2x +2xdy/dx+2y+4ydy/dx=0#
Divide everything by 2 and collect like terms
# color(white)("dddddddddddddddddddd") -> color(white)("d") xdy/dx+2ydy/dx = -x-y#
# color(white)("dddddddddddddddddddd") -> color(white)("d") dy/dx(x+2y) = -x-y#
# color(white)("dddddddddddddddddddd") ->color(white)("ddddddddd") dy/dx=(-x-y)/(x+2y)#
# color(white)("dddddddddddddddddddd") ->color(white)("ddddddddd") dy/dx=-(x+y)/(x+2y)#