Question

y=√1+√1+√x find the derivative

Absolutely no idea how to do this one.

Answer 1

sqrt(1) and sqrt(1) are just constants, the derivatives of constants = 0

Answer 2

1/(2sqrt(x))

Answer 3

and dy/dx sqrt(x)
=x^1/2
= 1/2(x)^-1/2 by the power rule

Answer 4

The answer in the back of the book is

1
------------------------------------------
8√x (√1+√x)(√1+√1+√x)

Answer 5

\[y=\sqrt{1+\sqrt{1+\sqrt{x}}}\]
is this what you meant ?

Answer 6

ah yes that's it lol sorry if that makes it completely different.

Answer 7

yes this makes it completely different

Answer 8

You need to know chain rule fo' sho'

Answer 9

Ya I'm just not sure which is the inside and which is out side since there's 3 terms.

Answer 10

\[y=(1+(1+x^\frac{1}{2})^\frac{1}{2})^\frac{1}{2}\]

Ok first of all it might help you to look this first:

\[y=f(g(h(x))) => y'=f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x)\]

Answer 11

\[y'=\frac{1}{2}(1+(1+x^\frac{1}{2})^\frac{1}{2})^{\frac{1}{2}-1} (1+(1+x^\frac{1}{2})^\frac{1}{2})'\]
But what is
\[(1+(1+x^\frac{1}{2})^\frac{1}{2})'\]
\[=(0+\frac{1}{2}(1+x^\frac{1}{2})^{\frac{1}{2}-1})(1+x^\frac{1}{2})'\]
But what is \[(1+x^\frac{1}{2})'\]
\[=0+\frac{1}{2}x^{\frac{1}{2}-1}\]
So we have....
\[y'=\frac{1}{2}(1+(1+x^\frac{1}{2})^\frac{1}{2})^\frac{-1}{2}(0+\frac{1}{2}(1+x^\frac{1}{2})^\frac{-1}{2})(0+\frac{1}{2}x^{\frac{-1}{2}})\]

Answer 12

lol that is nasty honestly

Answer 13

Ha ya I just hope that's not on my quiz tomorrow. Thanks for the help.

Answer 14

hey i got something better looking that you might understand better
you still there?

Answer 15

That made a lot more sense. I appreciate the time you took to do that.

Answer 16

I'm glad it looked better to you
I hope you understand what I did

Answer 17

I do indeed.