Find derivative of #x^2+2xy+2y^2=1#?

2 Answers
Mar 12, 2018

#dy/dx = -(x+ y)/(x + 2y)#

Explanation:

We have by implicit differentiation:

#2x + 2y + 2x(dy/dx) + 4y(dy/dx) = 0#

#dy/dx(2x + 4y) = -(2x + 2y)#

#dy/dx = -(2x + 2y)/(2x + 4y)#

#dy/dx = -(x+ y)/(x + 2y)#

Hopefully this helps!

Mar 12, 2018

#dy/dx= f'(x)=-(x+y)/(x+2y)#

Explanation:

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)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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)#

Tony B