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differentiate y=2(x+1)^3 by definition of differentiation..
help me?
 one year ago
 one year ago
differentiate y=2(x+1)^3 by definition of differentiation.. help me?
 one year ago
 one year ago

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privetekBest ResponseYou've already chosen the best response.0
chain rule  derivative of outside then derivative of inside: y'=2 (3(x+1)^(31)) *1 =6(x+1)^2
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.0
by the definition you have to use the limit as h approaches zero \[\lim_{h \rightarrow 0}\frac{ (x+h)x }{h }\]
 one year ago

sha0403Best ResponseYou've already chosen the best response.0
yes that is a formula... but how to find the answer using the formula? seriously i don't know..
 one year ago

JusaquikieBest ResponseYou've already chosen the best response.0
sorry i'm not sure, the way i tried didn't get me the right answer
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
Warning: I'm going to do this directly. \(y=2(x+1)^3\) \(y+\delta y=2(x+1+\delta x)^3\) Now, \(\delta y=2(x+1)^3 2(x+1+\delta x)^3 \) Note: \(x^3y^3=(xy)(x^2+y^2+xy)\) So, \(\delta y=2[(x+1(x+1+\delta x) ][(x+1)^2+(x+1+\delta x)^2+(x+1)(x+1+\delta x) \)] \(\delta y=2(\delta x)[(x+1)^2+(x+1+\delta x)^2+(x+1)(x+1+\delta x) \)] Dividing by \(\delta x\), \(\frac{\delta y}{\delta x}=2[(x+1)^2+(x+1+\delta x)^2+(x+1)(x+1+\delta x) \)] taking the limit, \( \delta x \rightarrow 0\) \[\lim_{\delta x \rightarrow 0}{\frac{\delta y}{\delta x}}=\lim_{\delta x \rightarrow 0}{2[(x+1)^2+(x+1+\delta x)^2+(x+1)(x+1+\delta x) ]}\] Thus, \[\frac{d y}{d x}=2[(x+1)^2+(x+1)^2+(x+1)(x+1) ]\] \[\frac{d y}{d x}=6(x+1)^2\] voila! Though I cannot imagine what kind of person would ask you to do this...
 one year ago

sha0403Best ResponseYou've already chosen the best response.0
hehe my lecturer ask to do this for my assignment..btw thanks for your help..i get it now.... ;)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
LOL okaaaay... You're welcome :)
 one year ago
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