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)
Find the derivative of f(x) = x^2 ln (x^2) using the Product Rule, and simplify.
3 answers
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)
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