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sha0403
Group Title
differentiate y=2(x+1)^3 by definition of differentiation..
help me?
 2 years ago
 2 years ago
sha0403 Group Title
differentiate y=2(x+1)^3 by definition of differentiation.. help me?
 2 years ago
 2 years ago

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privetek Group TitleBest 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
 2 years ago

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

sha0403 Group TitleBest 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..
 2 years ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
please help me?
 2 years ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
sorry i'm not sure, the way i tried didn't get me the right answer
 2 years ago

sha0403 Group TitleBest ResponseYou've already chosen the best response.0
oo its ok..thanks
 2 years ago

Shadowys Group TitleBest 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...
 2 years ago

sha0403 Group TitleBest 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.... ;)
 2 years ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
LOL okaaaay... You're welcome :)
 2 years ago
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