# 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)