Finding dy

[Angela123]

How do I find dy of (x+1)^(2/3) ? And then when x=-2 and dx=.0002 ?

 

[Deleted member 4993]

How do I find dy of (x+1)^(2/3) ? And then when x=-2 and dx=.0002 ?

Please post the COMPLETE and EXACT problem.

I assume you re given,

y = (x+1)[sup:292cpalh]2/3[/sup:292cpalh] and you need to find ?y at x = -2 and ?x = 0.0002

We know

?y = f(x+?x) - f(x)

Please show us your work indicating exactly where you are stuck - so that we know where to begin to help you.

 

[BigGlenntheHeavy]

$\displaystyle y \ = \ f(x ) \ = \ (x+1)^{2/3}$

$\displaystyle y' \ = \ f'(x ) \ = \ \frac{dy}{dx} \ = \ \frac{2}{3}(x+1)^{-1/3} \ = \ \frac{2}{3(x+1)^{1/3}}.$

$\displaystyle Hence \ dy \ = \ \frac{2dx}{3(x+1)^{1/3}}$

$\displaystyle Now \ plug \ in \ your \ values, \ and \ you're \ done.$

 

[Angela123]

I messed up. The whole problem is:

((x+1)^(2/3))-3 when x=-2 and dx=.0002

 

[Deleted member 4993]

I messed up. The whole problem is:

((x+1)^(2/3))-3 when x=-2 and dx=.0002

Follow the same process....