What is the dervative of ln(small 3 square root of (x^2)-3x+1)?

Is it (2x-3)/3(ln(x^2)-3x+1)^(2/3)?

To find the derivative of ln(small 3 square root of (x^2)-3x+1), we can make use of the chain rule. The chain rule states that if we have a function of the form f(g(x)), then the derivative is given by f'(g(x)) multiplied by g'(x).

In this case, let's define the function f(u) = ln(u), where u = small 3 square root of (x^2)-3x+1. The derivative of f(u) with respect to u is 1/u.

Next, we need to find the derivative of the function g(x) = small 3 square root of (x^2)-3x+1. We simplify this expression a bit first.

We have g(x) = (x^2 - 3x + 1)^(1/3).

Now, to find the derivative of g(x), we can use the power rule, which states that the derivative of x^n is n*x^(n-1).

Applying the power rule, we get:

g'(x) = (1/3)*(x^2 - 3x + 1)^(-2/3) * (2x - 3). Here, we have used the derivative of (x^2 - 3x + 1) as 2x - 3.

Now, we can apply the chain rule by multiplying the derivatives:

f'(g(x)) = 1/(small 3 square root of (x^2)-3x+1) * ((1/3)*(x^2 - 3x + 1)^(-2/3) * (2x - 3)).

Simplifying this expression, we get:

f'(g(x)) = (2x - 3)/(3*(small 3 square root of (x^2)-3x+1) * (x^2 - 3x + 1)^(2/3)).

Therefore, the derivative of ln(small 3 square root of (x^2)-3x+1) is indeed (2x - 3)/(3*(small 3 square root of (x^2)-3x+1) * (x^2 - 3x + 1)^(2/3)).