Find the derivative of f(x) = x^2 ln (x^2) using the Product Rule, and simplify.
Let x^2 = u and ln(x^2) = 2 ln x = v
f'(x) = u dv/dx + v du/dx
f'(x) = x^2*(2/x) + 2x *2 ln x)
= 2x + 4 x ln x = 2 x (ln x + 1)
y = (x^2) ln(x^2)
but ln(x^2) = 2 ln x
so
y = 2(x^2)ln x
dy/dx = 2(x^2) dy/dx (ln(x)) + 2 ln(x)dy/dx(x^2)
= 2 x^2 (1/x) + 2 ln x (2 x)
=2x +4x ln x
=2x(1+2lnx)
My last equation was wrong. Damon is right
To find the derivative of f(x) = x^2 ln(x^2) using the Product Rule, we need to differentiate both the first and second terms separately.
The Product Rule states that if we have two functions u(x) and v(x), the derivative of their product is given by the formula:
(d/dx)(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
Let's apply the Product Rule to f(x) = x^2 ln(x^2):
First, we need to identify the two functions:
u(x) = x^2
v(x) = ln(x^2)
Now, we differentiate these functions separately:
u'(x) = 2x (using the power rule)
v'(x) = 1 / (x^2) * 2x (using the chain rule)
Now, we plug these values back into the formula:
(d/dx)(x^2 ln(x^2)) = u'(x) * v(x) + u(x) * v'(x)
= (2x) * ln(x^2) + (x^2) * (1 / (x^2) * 2x)
= 2x ln(x^2) + 2x
So, the simplified derivative of f(x) = x^2 ln(x^2) is: 2x ln(x^2) + 2x.