We have the function f(x) = 2x^2 * ln x to differentiate.
Use the product rule:
f'(x) = (2x^2)' * ln x + 2x^2 * (ln x)'
f'(x) = 4x*ln x + 2x^2*(1/x)
f'(x) = 4x*ln x + 2x
The derivative of 2x^2 * ln x is 2x(ln (x^2) + 1)
We'll calculate the first derivative of f(x).
We'll put f(x) = y
We'll differentiate with respect to x:
dy/dx =(d/dx) (2x^2 * ln x)
dy/dx =[(d/dx) (2x^2)] * ln x + 2x^2*(d/dx)ln x)
dy/dx = 4xlnx + 2x^2/x
We'll simplify and we'll get:
dy/dx = 4xlnx + 2x
We'll factorize by 2x:
dy/dx = 2x(2lnx + 1)
dy/dx = 2x(ln x^2 + lne)
dy/dx = 2x ln(e*x^2)