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Finding dy
- Thread starter Angela123
- Start date
D
Deleted member 4993
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Angela123 said: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
Senior Member
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- Mar 8, 2009
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- 1,577
\(\displaystyle y \ = \ f(x ) \ = \ (x+1)^{2/3}\)
\(\displaystyle y' \ = \ f'(x ) \ = \ \frac{dy}{dx} \ = \ \frac{2}{3}(x+1)^{-1/3} \ = \ \frac{2}{3(x+1)^{1/3}}.\)
\(\displaystyle Hence \ dy \ = \ \frac{2dx}{3(x+1)^{1/3}}\)
\(\displaystyle Now \ plug \ in \ your \ values, \ and \ you're \ done.\)
\(\displaystyle y' \ = \ f'(x ) \ = \ \frac{dy}{dx} \ = \ \frac{2}{3}(x+1)^{-1/3} \ = \ \frac{2}{3(x+1)^{1/3}}.\)
\(\displaystyle Hence \ dy \ = \ \frac{2dx}{3(x+1)^{1/3}}\)
\(\displaystyle Now \ plug \ in \ your \ values, \ and \ you're \ done.\)
D
Deleted member 4993
Guest
Angela123 said:I messed up. The whole problem is:
((x+1)^(2/3))-3 when x=-2 and dx=.0002
Follow the same process....