How do you implicitly differentiate #7=1-(y-x)/y^2#?

1 Answer
Steve M
Nov 12, 2016

# dy/dx = 1/(1+12y) #

Explanation:

The equation does not express #y# explicitly in terms of #x# as in #y=f(x)#, instead we have #g(y)=f(x)#, so when we differentiate we apply the chain rule so that we differentiate #g(y)# wrt #y# rather than #x#, as in:

# g(y) = f(x) #
# :. d/dx g(y) = d/dx f(x) #
# :. dy/dx d/dy g(y) = f'(x) #
# :. g'(y) dy/dx= f'(x) #

So for # 7=1-(y-x)/y^2 # we have:

# -6 = (y-x)/y^2#
# :. -6y^2 = y-x #
# :. -12ydy/dx = dy/dx-1 #
# :. dy/dx+12ydy/dx = 1 #
# :. (1+12y)dy/dx = 1 #
# :. dy/dx = 1/(1+12y) #

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