What is the derivative
(tˆ2)ˆ(1/3)+2(tˆ5)ˆ(1/2)
3 answers
what did you get? These are simple power rule problems.
i got [2tˆ(1/3)]+[(10tˆ4)ˆ4
no, you gotta do a little work here.
(t^2)^(1/3) = t^(2/3)
(t^5)^(1/2) = t^5/2
so, you have
y = t^(2/3) + 2t^(5/2)
y' = 2/3 t^(-1/3) + 2(5/2) t^(3/2)
= 2/3 t^(-1/3) + 5t^(3/2)
you need to review your chain rule some more.
Just working it as-is, using the chain rule, you get
if y = u^n, y' = n u^(-1) u'
1/3 (t^2)^(-2/3) * (2t) + 2 (1/2)(t^5)^(-1/2) * (5t^4)
= 2/3 t t^(-4/3) + 5t^4 t^(-5/2)
= 2/3 t^(-1/3) + 5t^(3/2)
(t^2)^(1/3) = t^(2/3)
(t^5)^(1/2) = t^5/2
so, you have
y = t^(2/3) + 2t^(5/2)
y' = 2/3 t^(-1/3) + 2(5/2) t^(3/2)
= 2/3 t^(-1/3) + 5t^(3/2)
you need to review your chain rule some more.
Just working it as-is, using the chain rule, you get
if y = u^n, y' = n u^(-1) u'
1/3 (t^2)^(-2/3) * (2t) + 2 (1/2)(t^5)^(-1/2) * (5t^4)
= 2/3 t t^(-4/3) + 5t^4 t^(-5/2)
= 2/3 t^(-1/3) + 5t^(3/2)
5 answers