# x^2/(e^(x-y))=C #
# :. x^2=Ce^(x-y) #
# :. x^2=Ce^xe^-y #
Differentiating implicitly involves differentiating everything wrt #x# and using the chain rule and product rule:
So, # d/dx(x^2)=d/dx(Ce^xe^-y) #
# :. 2x=Cd/dx(e^xe^-y) #
# :. 2x=C(e^xd/dxe^-y+e^-yd/dxe^x) # using the product rule
# :. 2x=C(e^xd/dye^-ydy/dx+e^-ye^x) # using the chain rule
# :. 2x=C(-e^xe^-ydy/dx+e^-ye^x) #
# :. 2x=x^2/(e^(x-y))(-e^xe^-ydy/dx+e^-ye^x) #
We don't need an explicit expression for #dy/dx# so don't attempt to simplify further. We can just substitute #x=-2# and #y=1#, to find #dy/dx# at that value.
So, at # (-2,1) # we have:
# :. 2x=x^2/(e^(x-y))(-e^xe^-ydy/dx+e^-ye^x) #
# :. 2(-2)=(-2)^2/(e^(-2-1))(-e^-2e^-1 dy/dx+e^-1e^-2)#
# :. -4=4/(e^-3)(-e^-3 dy/dx+e^-3)#
# :. -4=4(- dy/dx+1)#
# :. - dy/dx+1=-1#
# :. dy/dx=2#
So the tangent passes through #(-2,1)# and has slope #m=2# so we use the formula #y-y_1=m(x-x_1)# to get the equation:
# y-1=2(x--2) #
# :. y-1=2(x+2) #
# :. y-1=2x+4 #
# :. y=2x+5 #
