anonymous
  • anonymous
Find the derivative of the function. y=x/sqrt(x^2+1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
use uv rule
anonymous
  • anonymous
\[y=x/\sqrt{x ^{2}+1}\]
anonymous
  • anonymous
well I understand part of it

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anonymous
  • anonymous
use quotient rule
anonymous
  • anonymous
U-substitution, yo. Or whatever they call it nowadays.
anonymous
  • anonymous
then gemeraal power rule
anonymous
  • anonymous
general
anonymous
  • anonymous
Oh, wait, derivative, not integral. XD
anonymous
  • anonymous
i get to this answer
anonymous
  • anonymous
y=x/sqrt(x^2+1) Use Quenient and Chain rule
anonymous
  • anonymous
Alright, product rule of x*(x^2+1)^(-1/2), (x^2+1)^(-1/2)-(1/2)x(2x)(x^2+1)^(-3/2) Yeah.
anonymous
  • anonymous
\[y'=(x ^{2}+1)^{1/2} - x ^{2}(x ^{2}+1)^{-1/2}/x ^{2}+1\]
anonymous
  • anonymous
thqats the derivative now i have to figure out how to simplify it
anonymous
  • anonymous
is the answer
anonymous
  • anonymous
oh crud made a mistake (sqrt(x^2+1)) - x(2x/2sqrt((x^(2)+1)))/(sqrt(x^2+1))^(2)
anonymous
  • anonymous
that is the answer
anonymous
  • anonymous
because the final answer in the back of the book is \[1/\sqrt{(x ^{2}+1)^{3}}\]
anonymous
  • anonymous
thats the answer the book gives me
anonymous
  • anonymous
i know my first answer is correct but not simplified
anonymous
  • anonymous
simplify
anonymous
  • anonymous
if i understood how to simplify it I wouldnt be here
anonymous
  • anonymous
lol
anonymous
  • anonymous
fair enough :) i will help or try to my battery is about to die
anonymous
  • anonymous
( (sqrt(x^2+1)) - x(2x/2sqrt((x^(2)+1))) )/(sqrt(x^2+1))^(2) = ( (sqrt(x^2+1)) - (x/sqrt(x^(2)+1)) )/ (sqrt(x^2+1))^(2) = (sqrt(x^2+1)) / (sqrt(x^2+1))^(2) - ( (x/sqrt(x^(2)+1)) / (sqrt(x^2+1))^(2) ) = 1 / (sqrt(x^2+1))^(2) - ( (x/sqrt(x^(2)+1)) / (sqrt(x^2+1))^(2) )
anonymous
  • anonymous
Truly, simplification and algebraic manipulation are the difficult parts of calculus; not the class' own namesake.
Mimi_x3
  • Mimi_x3
Lol, yes it is in the quotient rule, thats why i hate it. xD
anonymous
  • anonymous
somehow it simplifies to
anonymous
  • anonymous
1 / (sqrt(x^2+1))^(2) - ( (x/sqrt(x^(2)+1)) / (sqrt(x^2+1))^(2) ) = 1 / (sqrt(x^2+1))^(2) - x(sqrt(x^2+1))^(2) )/sqrt(x^(2)+1) = 1 / (sqrt(x^2+1))^(2) - x(sqrt(x^2+1))/1
anonymous
  • anonymous
1 / (sqrt(x^2+1)) - x(sqrt(x^2+1)) sorry made a mistake
anonymous
  • anonymous
\[(x ^{2}+1)^{-3/2}(x ^{2}+1)/(x^2=1)\]
anonymous
  • anonymous
/x^2+1) oops
anonymous
  • anonymous
But yeah try multiplying both the top and the bottom by the conjugate: (x(sqrt(x^2+1))^(2) + 1) but as far as I would go to simplify this would be it it ( x(sqrt(x^2+1))^(2) - 1 )/(sqrt(x^2+1))
anonymous
  • anonymous
for final answer power in denominator should be 3/2
anonymous
  • anonymous
\[(\sqrt{x ^{2}+1} - x^2/\sqrt{x^2+1})/(x^2+1)\]
anonymous
  • anonymous
\[(x^2+1-x^2)/\sqrt{x^2+1}/x^2+1\]
anonymous
  • anonymous
\[1/(x^2+1)(\sqrt{x^2+1})\]
anonymous
  • anonymous
would \[(x^2+1)(\sqrt{x^2+1})= \sqrt{(x^2+1)^3}\]
anonymous
  • anonymous
because the final answer should be 1/sqrt(x+1)^3
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=y%3Dx%2Fsqrt%28x%5E2%2B1%29+find+derivative
anonymous
  • anonymous
i understand that marina same thing i have there
anonymous
  • anonymous
(x+1)^2/3 = sqrt((x+1)^2)
anonymous
  • anonymous
^3 oops
Diyadiya
  • Diyadiya
you're right y\[(x^2+1) \sqrt{x^2+1}= (x^2+1)(x^2+1)^{1/2}= (x^2+1)^{3/2}\]
anonymous
  • anonymous
i do believe \[(x^2+1)(\sqrt{x^2+1}) = \sqrt{(x^2+1)^3}\]
anonymous
  • anonymous
ok :) then the way i did that is right
Diyadiya
  • Diyadiya
Right!

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