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If , #y=sqrt(x/a)-sqrt(a/x)# So, Prove that ?? #2xy(dy/dx)= x/a-a/x#

If , #y=sqrt(x/a)-sqrt(a/x)# So, Prove that #?#
#2xydy/dx= x/a-a/x#

If , #y=sqrt(x/a)-sqrt(a/x)# So, Prove that #?#
#2xydy/dx= x/a-a/x#

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Steve M Share
Jun 17, 2018

We seek to show that:

#2xydy/dx= x/a-a/x# where #y=sqrt(x/a)-sqrt(a/x)#

Squaring the expression, and then expanding, we get:

# y^2 = (sqrt(x/a)-sqrt(a/x))^2 #
# \ \ \ \= (sqrt(x/a))^2 - 2(sqrt(x/a))(sqrt(a/x)) + (sqrt(a/x))^2#
# \ \ \ \= x/a - 2 + a/x #

Differentiating Implicitly, we get

# 2y dy/dx = 1/a -a/x^2 #

Finally, multiplying by #x# we get the desired result:

# 2xy dy/dx = x/a -a/x \ \ \ # QED

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Jun 17, 2018

Answer:

Kindly see a Proof in Explanation.

Explanation:

We have, #y=sqrt(x/a)-sqrt(a/x)#.

Diff.ing w.r.t. #x, dy/dx=1/sqrta*(1/2x^(-1/2))-sqrta*(-1/2x^(-3/2))#.

#:. dy/dx=1/(2sqrt(ax))+sqrta/(2xsqrtx)#.

Multiplying by #2x, 2xdy/dx=sqrt(x/a)+sqrt(a/x)#.

Again multiplying by #y, 2xydy/dx=y(sqrt(x/a)+sqrt(a/x))#,

# i.e., 2xydy/dx=(sqrt(x/a)-sqrt(a/x))(sqrt(x/a)+sqrt(a/x))#.

# rArr 2xydy/dx=x/a-a/x#, as desired!

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Jun 17, 2018

Given: #y=sqrt(x/a)-sqrt(a/x)#

Differentiating:

#dy/dx = (sqrt(a/x) + sqrt(x/a))/(2 x)#

Multiply both sides by2xy:

#2xydy/dx = 2xy(sqrt(a/x) + sqrt(x/a))/(2 x)#

#(2x)/(2x)# becomes 1:

#2xydy/dx = y(sqrt(a/x) + sqrt(x/a))#

Substitute #y=sqrt(x/a)-sqrt(a/x)# on the right side only:

#2xydy/dx = (sqrt(x/a)-sqrt(a/x))(sqrt(a/x) + sqrt(x/a))#

Please observe that the right side is the pattern #(a - b)(a+b) = a^2-b^2# where #a = sqrt(x/a)# and #b = sqrt(a/x)#

#2xydy/dx = x/a-a/x" Q.E.D."#

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