How to find the derivative of y=r/(sqrt(r^2 + 1)?

Question

the answer is y'= (r^2 + 1)^-3/2 , I tried solving it, still I get wrong answer.

Please help.

Answer 1

... y = r (r² + 1)^(-1/2)

or dy/dr = r d/dr [(r² + 1)^(-1/2)] + (r² + 1)^(-1/2) d/dr [r]

or dy/dr = (-1/2) r (r² + 1)^(-3/2) d/dr [r²] + (r² + 1)^(-1/2)

or dy/dr = (-1/2) r (r² + 1)^(-3/2) [2r] + (r² + 1)^(-1/2)

or dy/dr = - r² / (r² + 1)^(3/2) + 1/ (r² + 1)^(1/2)

or dy/dr = [ - r² + (r² + 1) ] / (r² + 1)^(3/2)

or dy/dr = 1 / (r² + 1)^(3/2)

Answer 2

y = r/√(r²+1)

= r(r²+1)^-½



Use the Product Rule:

y' = r * [-½(2r)(r²+1)^(-3/2)] + (r²+1)^-½

= r * [-r/(r²+1)^(3/2)] + 1/(r²+1)^½

= -r²/(r²+1)^(3/2) + (r²+1)/(r²+1)^(3/2)

= (r²+1-r²) / (r²+1)^(3/2)

= 1/(r²+1)^(3/2)

= (r²+1)^(-3/2)

Answer 3

y = [1* sqrt(r^2 + 1) - r * 1/2sqrt(r^2 + 1)* 2r] / [sqrt(r^2 + 1)]^2 =

[(r^2 + 1 - r^2)/sqrt(r^2 + 1)]/[sqrt(r^2 + 1)]^2 =

(r^2 + 1)^( -3/2)

Answer 4

y = r / ( r ² + 1 )^(1/2)

dy/dx by Quotient Rule :-



( r ² + 1 )^(1/2) - r (1/2) ( r ² + 1 )^(-1/2) ( 2 r )

---------------------------------------------------------

( r ² + 1 )



( r ² + 1 )^(1/2) - r ² ( r ² + 1 )^(-1/2)

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( r ² + 1 )



( r ² + 1 )^(-1/2) [ ( r ² + 1 ) - r ² ]

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r ² + 1



1

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( r ² + 1 )^(3/2