How do you use implicit differentiation to find dy/dx given #2^sqrt(xy)=x#?

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Answer 1 · 2017-01-17T00:09:50

#(dy)/(dx) = lnx/(xln2)^2 ( 2-lnx)#

Differentiate both members of the equation with respect to #x# keeping in mind that:

#d/(dx) f(y(x)) = (df)/(dy)*(dy)/(dx)#

#d/(dx) 2^sqrt(xy) = d/(dx) x#

#ln2 ((y+xy')/(2sqrt(xy)) )2^sqrt(xy) = 1#

Substitute #2^sqrt(xy) = x# from the original equation:

#ln2 ((y+xy')/(2sqrt(xy)) )x= 1#

Solving for #y'#:

#(y+xy') = 2/ln2 sqrt(xy)/x#

#y' =-y/x + 2/ln2 sqrt(xy)/x^2#

Now note that:

#2^sqrt(xy) = x => sqrt(xy) = lnx/ln2#

and so:

#xy =ln^2x/ln^2 2#

#y =1/xln^2x/ln^2 2#

Substituting in the expression of #y'# we have:

#y' =-1/(x^2) ln^2x/ln^2 2 + 2/ln^2 2 lnx/x^2= lnx/(xln2)^2 ( 2-lnx)#

You can also first make #y# explicit from the original equation and then differentiate:

#y =1/xln^2x/ln^2 2#

#(dy)/(dx) = (-1/x^2)ln^2x/ln^2 2 +1/x 1/ln^2 2 2lnx/x= lnx/(xln2)^2 ( 2-lnx)#

Obtaining naturally the same result.

Answer 2 · 2017-01-17T01:09:19

x and y > 0. #y'=1/x(2/ln2sqrt(y/x)-y)#.

graph{x-2^sqrt(xy)=0x^2 [-39.94, 39.94, -19.96, 19.98]} To make the function real, #x >0#, giving #y >0#.

#y to oo#, as #x to 0_+#.

Using logarithmic differentiation,

#ln 2 sqrt(xy)'=(ln x)'#, giving

#1/2(y'sqrt(x/y)+sqrt(y/x))=1/(xln2)#. So,

#y'=(2/(xln2)-sqrt(y/x))sqrt(y/x)#

#=1/x(2/ln2sqrt(y/x)-y)#