I need to find the dy/dx of x^2ln(x^2)I tried to do it the way I thought, but it did not come out right.

The product rule states that if you have two functions, u(x) and v(x), the derivative of their product is given by:

(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)

In this case, let u(x) = x^2 and v(x) = ln(x^2). Now we can proceed step by step:

Step 1: Find u'(x) - the derivative of u(x):
u(x) = x^2
Applying the power rule of differentiation, we get:
u'(x) = 2x

Step 2: Find v'(x) - the derivative of v(x):
v(x) = ln(x^2)
To differentiate ln(x^2), we can use the chain rule. Let's define a new function, h(x) = x^2, and rewrite v(x) as v(h(x)) = ln(h(x)).
Applying the chain rule, we get:
v'(x) = (d/dx)ln(h(x)) = (1/h(x)) * h'(x)

To find h'(x), we differentiate x^2 using the power rule:
h(x) = x^2
h'(x) = 2x

Now we substitute h'(x) and h(x) back into v'(x):
v'(x) = (1/h(x)) * h'(x)
v'(x) = (1/x^2) * (2x)
v'(x) = 2/x

Step 3: Apply the product rule to find the derivative of f(x):
(d/dx)(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)
Substituting the values we found earlier:
(d/dx)(x^2ln(x^2)) = 2x * ln(x^2) + x^2 * (2/x)

Simplifying the expression:
2x * ln(x^2) + 2x = 2x(ln(x^2) + 1)

So, the derivative of f(x) = x^2ln(x^2) is 2x(ln(x^2) + 1).