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

1. privetek Group Title

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

2. Jusaquikie Group Title

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

3. sha0403 Group Title

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

4. sha0403 Group Title

5. Jusaquikie Group Title

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

6. sha0403 Group Title

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...

8. sha0403 Group Title

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