How do you find the derivative of #y = lnx^2#?

2 Answers
Jun 29, 2015

Answer:

Either use the chain rule and #d/dx(lnu) = 1/u (du)/dx#, or use properties of the logarithm to rewrite in simpler form.

Explanation:

#y = lnx^2#

I assume that we are using correct notation and the function here is #f(x) = ln(x^2)#
(If we meant the square of the #ln# we would have to write #(lnx)^2# or, perhaps #ln^2x#.)

Chain Rule Solution

#d/dx(lnx^2) = 1/x^2 * d/dx(x^2) = 1/x^2 * 2x = 2/x#

Rewrite Solution

Use #lna^r = rlna#, to get:

#d/dx(lnx^2) = d/dx(2lnx) = 2 d/dx(lnx) = 2(1/x) = 2/x#

Jul 1, 2015

Alternatively, if you have some free time, you can do some manipulation to this and get an idea of what it means to implicitly differentiate.

#e^y = x^2#
Now we have #z = z(y(x)) = e^y = x^2#
where #z(y) = e^y# and #y(x) = x^2#.

And #(dz)/(dx) = ((dz)/(dy)) ((dy)/(dx))#

Thus:
#(dz)/(dx) = ((d[e^y])/(dy)) ((dy)/(dx)) = e^y (dy)/(dx) = 2x#

So now we just get:

#(dy)/(dx) = (2x)/(e^y) = (2x)/(x^2) = 2/x#