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.
dy/dx= 2xlnx^2 + x^2/x^2 *2x
reduce it.
If I did it correctly I got 4x ln (x^2).
I got 2x (1+lnx^2)
You are right
To find the derivative of the function f(x) = x^2ln(x^2), you can use the product rule and the chain rule.
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).