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

1. anonymous

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

2. Jusaquikie

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

3. anonymous

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

4. anonymous

5. Jusaquikie

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

6. anonymous

oo its ok..thanks

7. anonymous

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...

8. anonymous

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

9. anonymous

LOL okaaaay... You're welcome :)