how would u find this derivative?

Question

t(4-t)^1/2 t< 3

Answer 1

use the product rule, with the first term as t and the second as (4-t)^1/2
product rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term. Don't forget to use the chain rule along the way.

That is
xy would have a derivate of yx' =xy'

So here
[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]
(4-t)^1/2 + -t/2(4-t)^ -1/2

Answer 2

I believe you want to use the chain rule in addition the product rule.

If you set f(t)=t and g(t)=(4-t)^1/2, then f'(t)=1, and we need to do some further expansion to find g'(t).

For (4-t)^1/2, let h(t)=x^1/2 and k(t)=4-t. Then, h'(t)=1/2x^-1/2 and k'(t)=1. The chain rule says

F'(t)=h'(k(t))k'(t)

So we take the h'(t) function and substitute k(t) for t

1/2(4-t)^1/2

and multiply this by k'(t)

(1/2(4-t)^1/2)(1)=1/2(4-t)^1/2

Now BACK to our original problem, if g(t)=(4-t)^1/2, then g'(t)=1/2(4-t)^1/2

And now we can use the product rule

dy/dt=f'(t)g(t)+f(t)g'(t)
=(1)((4-t)^1/2)+(t)(1/2(4-t)^1/2)
=(4-t)^1/2 + 1/2t(4-t)^1/2