how do you find the derivative (dy/dt) of this problem??

Question

dy/dt sq.rt.(1+ sq.rt(t))

it looks a whole lot easier if you write it down how it says too.

pleeeaassee help thank youuu

Answer 1

If you rewrite the sqrt as an exponent it is easier.

(1 + t^(1/2))^(1/2)
This is the same as 1^(1/2) + t^(1/4)
This is the same as 1 + t^(1/4)

Now just use the rules for takeing the derivative and you get
(1/4)t^(-3/4)
or
1/4t^(3/4)

Hope that Helped.....

Answer 2

differentiate the brakets as a whole unit of power
1/2 then multiply this with the derivative of the inside of the brackets you get.
1/2sqrt(t)) *(1+sqrt(t))

Answer 3

dy/dt = (1/2)(1/(1+sqrt(t))^.5 * 1/t^.5
dy/dt = 1/(2t^.5(1+t^.5)^.5)

Answer 4

i assume the problem is y=sq rt (1+sq rt (t) ) rewrite it as:
y=(1+t^(.5) ) ^(.5)
then chain rule
dy/dt =.5(1+t^(.5))^(-.5) * .5(t)^(-.5)