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

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chain rule - derivative of outside then derivative of inside: y'=2 (3(x+1)^(3-1)) *1 =6(x+1)^2
by the definition you have to use the limit as h approaches zero \[\lim_{h \rightarrow 0}\frac{ (x+h)-x }{h }\]
yes that is a formula... but how to find the answer using the formula? seriously i don't know..

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please help me?
sorry i'm not sure, the way i tried didn't get me the right answer
oo its ok..thanks
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: \(x^3-y^3=(x-y)(x^2+y^2+xy)\) 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...
hehe my lecturer ask to do this for my assignment..btw thanks for your help..i get it now.... ;)
LOL okaaaay... You're welcome :)

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