anonymous
  • anonymous
y=√1+√1+√x find the derivative Absolutely no idea how to do this one.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
sqrt(1) and sqrt(1) are just constants, the derivatives of constants = 0
bahrom7893
  • bahrom7893
1/(2sqrt(x))
anonymous
  • anonymous
and dy/dx sqrt(x) =x^1/2 = 1/2(x)^-1/2 by the power rule

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anonymous
  • anonymous
The answer in the back of the book is 1 ------------------------------------------ 8√x (√1+√x)(√1+√1+√x)
myininaya
  • myininaya
\[y=\sqrt{1+\sqrt{1+\sqrt{x}}}\] is this what you meant ?
anonymous
  • anonymous
ah yes that's it lol sorry if that makes it completely different.
myininaya
  • myininaya
yes this makes it completely different
myininaya
  • myininaya
You need to know chain rule fo' sho'
anonymous
  • anonymous
Ya I'm just not sure which is the inside and which is out side since there's 3 terms.
myininaya
  • myininaya
\[y=(1+(1+x^\frac{1}{2})^\frac{1}{2})^\frac{1}{2}\] Ok first of all it might help you to look this first: \[y=f(g(h(x))) => y'=f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x)\]
myininaya
  • myininaya
\[y'=\frac{1}{2}(1+(1+x^\frac{1}{2})^\frac{1}{2})^{\frac{1}{2}-1} (1+(1+x^\frac{1}{2})^\frac{1}{2})'\] 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.... \[y'=\frac{1}{2}(1+(1+x^\frac{1}{2})^\frac{1}{2})^\frac{-1}{2}(0+\frac{1}{2}(1+x^\frac{1}{2})^\frac{-1}{2})(0+\frac{1}{2}x^{\frac{-1}{2}})\]
myininaya
  • myininaya
lol that is nasty honestly
anonymous
  • anonymous
Ha ya I just hope that's not on my quiz tomorrow. Thanks for the help.
myininaya
  • myininaya
hey i got something better looking that you might understand better you still there?
myininaya
  • myininaya
anonymous
  • anonymous
That made a lot more sense. I appreciate the time you took to do that.
myininaya
  • myininaya
I'm glad it looked better to you I hope you understand what I did
anonymous
  • anonymous
I do indeed.

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