Question

derivative:


1. g(t)=√√t+1 +1

Answer 1

you need to better clarify the function

Answer 2

\[g(t) = \sqrt{\sqrt{t+1}}+1\]

Answer 3

the +1 is included inside the big sqrt :D

Answer 4

\[g(t) = \sqrt{\sqrt{t+1}+1}~~\]

Answer 5

yep thats it

Answer 6

its best to work with powers, sqrt = ^1/2

other than that, it looks like power and chain rules to apply

Answer 7

have you covered the derivative rules yet, or is this one of thos first principles thing with the limit of a slope?

Answer 8

after ^1/2 apply chain rule?

Answer 9

\[((t+1)^{1/2}+1)^{1/2}\]
we have a function in a function, so chain rule applies:\[f(g)=f'(g)~g'\]

Answer 10

let f = a^(1/2) f' = a'/[2a^(1/2)]

but a = (t+1)^(1/2) + 1, so a' = 1/[2(t+1)^(1/2)]

agreed?

Answer 11

\[1/2(\sqrt{t+1}+1)x((t+1)^{1/2}) \]

Answer 12

\[((t+1)^{1/2}+1)^{1/2}\]
\[\frac12((t+1)^{1/2}+1)^{-1/2}~*~((t+1)^{1/2}+1)'\]
\[\frac12((t+1)^{1/2}+1)^{-1/2}~*~\frac12(t+1)^{-1/2}\]
\[\frac14((t+1)^{1/2}+1)^{-1/2}(t+1)^{-1/2}\]
\[\frac14[(t+1)^{1/2}+1)(t+1)]^{-1/2}\]

Answer 13

or written in sqrts:\[\frac{1}{4\sqrt{(\sqrt{t+1}+1)(t+1)}}\]

Answer 14

can you help me with another one?