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July 31, 2014

July 31, 2014

Posted by **Mel <3** on Sunday, November 30, 2008 at 6:58pm.

- Calculus -
**drwls**, Sunday, November 30, 2008 at 7:37pmLet 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)

- Calculus -
**Damon**, Sunday, November 30, 2008 at 7:41pmy = (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)

- Calculus -
**drwls**, Sunday, November 30, 2008 at 7:51pmMy last equation was wrong. Damon is right

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