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Mathematics 16 Online
OpenStudy (anonymous):

Find the derivative

OpenStudy (anonymous):

\[t^{3/2} (2+\sqrt{t})\]

OpenStudy (turingtest):

do you know the derivative of t^(2/3) ?

OpenStudy (turingtest):

t^(3/2) excuse me....

OpenStudy (anonymous):

3/2t^3?

OpenStudy (anonymous):

do you know how to apply the product rule and power rule?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

i just have trouble finding the derivative for the square root. i got -1/2x, i'm not sure if that's right

OpenStudy (anonymous):

almost right.

OpenStudy (anonymous):

sqroot t = t^(1/2)

OpenStudy (anonymous):

that's the derivative? isn't that the same as square root of t?

OpenStudy (anonymous):

but with an exponent?

OpenStudy (anonymous):

yea, so apply the power rule to find the derivative

OpenStudy (anonymous):

f * g' + f' * g -> t^3/2 * t^1/2 + 3/2t^3 * ? are you sure that is the derivative of square root of t?

OpenStudy (anonymous):

no, i was just showing you that square root t can be rewritten like that. apply the product rule to find the derivative of sq root t

OpenStudy (anonymous):

but that's what i have problems with? i got 1/2x as the derivative of square root of t

OpenStudy (anonymous):

dx^1/2/dx = (x^[-1/2])/2

OpenStudy (anonymous):

= 1/(2x^[1/2])

OpenStudy (anonymous):

|dw:1349841522489:dw|

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