anonymous
  • anonymous
Derivatives, Chain Rule, Numero 3 (Please help!): \(\ G(x)=(3x-2)^{10} (5x^2-x+1)^{12} \)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
10(3x-2)^9*3*(5x.....)+(3x...)*12(5x..)^11.(10x-1)
anonymous
  • anonymous
So take the power rule and then use the product rule?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
Cool! Thank you @studentphy
anonymous
  • anonymous
Most WC
zepdrix
  • zepdrix
Product rule THEN power rule on each term that requires differentiation :O Just thought I'd mention that in case there was any confusion <:o
anonymous
  • anonymous
I just got stuck on that problem again. Thanks for the clarification, yet again, @zepdrix!!
anonymous
  • anonymous
d/dx(u*V)=keep u constant * derivative of v (which is use power rule)+keep v constant *derivative of U(which is first use power rule then product) hope it clarifies.
anonymous
  • anonymous
Wait a minute.... How can you differentiate the two terms with exponents when using the product rule before power...? Could someone show me step by step please
zepdrix
  • zepdrix
|dw:1352354812852:dw|
anonymous
  • anonymous
\(\ \Huge \text{ Eureka!} \) \(\ \large \text{Thanks for clarifying, that makes sense!} \)
anonymous
  • anonymous
suppose u have x^n *2Y then d/dx= nx?^n-1* 2Y+ 2(which is after differentiating y )* X^n
zepdrix
  • zepdrix
Grrr you and your fancy latex! >:3
anonymous
  • anonymous
\(\ \text{Hahaha zepdrix! I actually wasn't all that good when I joined OS 6 month ago.}\) \(\ \text{I've practiced a lot, and look at the latex of other people - which is how I learned.} \) \(\ \text{I'm still learning new things. Practice makes perfect!} \) \(\ \text{ I recently learned how to write text in LATEX. You can be fancy too, with practice!}\)

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