How do you use implicit differentiation to find dy/dx given #ln(x-2)=ln(2y+1)#?
1 Answer
# dy/dx=1/2 #
Explanation:
We don't actually need to use implicit differentiation (at least in the logarithmic form) due to the fact that:
# ln A = ln B iff A=B #
So if:
# ln(x-2) = ln(2y+1) iff x-2=2y+1 #
# :. 2y=x-3 #
# :. \ \ y=1/2x-3/3 #
And so if we differentiate we get;
# dy/dx=1/2 #
If you are not convinced by that or just just love Calculus then we can differentiate the initial equation implicitly:
# ln(x-2) = ln(2y+1) iff x-2=2y+1 #
# :. 1/(x-2) = 1/(2y+1)*2*dy/dx #
# :. 2/(2y+1) * dy/dx = 1/(x-2)#
# :. 2dy/dx = (2y+1)/(x-2)#
Now at a first glance this seems very different to the answer above, but is in fact the same, which we can show as follows:
Take Natural Logarithms of both sides:
# ln{2dy/dx} = ln {(2y+1)/(x-2)} #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ln (2y+1) - ln(x-2) \ \ \ # (rule of logs)
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 \ \ \ # (from the initial equation)
# :. 2dy/dx = e^0 #
# \ \ \ \ \ \ \ \ \ \ \ \ = 1 #
# \ \ \ \ \ \ \ dy/dx = 1/2 \ \ \ # (as above)
