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

2 Answers
Jun 29, 2015

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