Question

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

Answer 1

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`

Answer 2

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`