Please help with a calculus problem: Find the derivative of y=x^2 ln x^2?

Answer 1

According to the properties of the log function, ln(x^2) = 2 ln(x). So, y = 2x^2 ln(x). Considering the rule for differentiating a product, we get

y' = 2 [x^2 (ln(x)' + (x^2)' ln(x)] Since (ln(x)' = 1/x and (x^2)' = 2x, it follows that

y' = 2[x^2 * 1/x + 2x ln(x)] = 2[ x + 2x ln(x)] = 2x [1 + 2 ln(x)], whic is the same as y' = 2x [1 + ln(x^2)]

Answer 2

product rule of Calculus
a*b'+a'b

set up your value first

set a =x^2 and b=ln(x^2)

now find the derivative of each
a'=2x b'=(2*x)/(x^2)

place the value together and you get
x^2*(2*x)/(x^2)+2x*ln(x^2)

simplify
2x+2x*ln(x^2)

Answer 3

Product rule and chain rule:

2 x + 2 x Log[x^2]

Answer 4

2*x * lin(x^2) + x^2 ((2*x)/(x^2))

Product Rule is used.

Answer 5

y'=2x lnx^2+(2x/x^2) * x^2

=2xlnx^2+2x

=2x(1+lnx^2)

Answer 6

easy....
dy/dx=[(x^2)*(1/x^2)*2x ]+[2xln(x^2)],
so, by solving u will get-
dy/dx=2x[1 + lnx^2].