What's the derivative?

Question

Can someone help me find the derivative of

y=[ln(x^2)]^2

An answer is nice, but an explanation is better

Answer 1

Chain rule says that, if you have 3 "chained" functions:

f(g(h(h(x)))

Then the derivative is computed by taking the derivative of each function in turn, starting at the outside and working your way in:

f'(g(h(x)) × g'(h(x)) × h'(x)

In this case:

f(x) = x²
g(x) = ln(x)
h(x) = x²

So that f(g(h(x))) = [ln(x²)]² (try plugging them in to verify).

Then the derivatives are:

f'(x) = 2x
g'(x) = 1/x
h'(x) = 2x

So:

[[ln(x²)]²]' = 2[ln(x²)] × [1/x²] × [2x] = 4[ln(x²)]/x = 8 ln(x)/x

If I didn't mess up. (That last step is optional, and just uses the power law of logs to make it look a little nicer.)

Answer 2

Your function is in the form u^2 which has derivative

2u (du/dx) by the power and chain rules. But since

u = ln(x^2), du/dx = 1/x^2 [d(x^2)/dx] = 2x/x^2 = 2/x.

Putting all these together we can write

y' = 2ln(x^2) d[ln(x^2)]/dx = 2ln(x^2) 1/x^2 d(x^2)/dx

= 2ln(x^2) (1/x^2) (2x) = [4ln(x^2)] / x

Answer 3

First use the power rule to get

dy/dx = 2 (ln(x^2)) d/dx(ln(x^2)).

Now find d/dx (ln(x^2)).

dy/dx = 2 (ln(x^2)) (1/(x^2)) dx(x^2)

Now find d/dx (x^2).

dy/dx = 2 (lnx^2)) (1/(x^2)) (2x)

dy/dx = 4/x (lnx^2)

Answer 4

Let u = x ²
y =( ln u ) ²
dy/du = 2 (ln u) (1/u)
du/dx = 2x

dy/dx = (2x) 2 (ln u) (1 / u)
dy/dx = (2x) 2 ((ln x ²) (1 / x²)
dy/dx = (4 / x) [ ln ( x ²) ]