Question

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

Answer 1

chain rule - derivative of outside then derivative of inside:
y'=2 (3(x+1)^(3-1)) *1
=6(x+1)^2

Answer 2

by the definition you have to use the limit as h approaches zero \[\lim_{h \rightarrow 0}\frac{ (x+h)-x }{h }\]

Answer 3

yes that is a formula... but how to find the answer using the formula? seriously i don't know..

Answer 4

please help me?

Answer 5

sorry i'm not sure, the way i tried didn't get me the right answer

Answer 6

oo its ok..thanks

Answer 7

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...

Answer 8

hehe my lecturer ask to do this for my assignment..btw thanks for your help..i get it now.... ;)

Answer 9

LOL okaaaay...
You're welcome :)