How do you find the differential #dy# of the function #y=sqrtx+1/sqrtx#?

1 Answer
Sep 4, 2017

Note that #1/sqrtx = x^(-1/2)#, and #sqrt x = x^(1/2)#, then use the power rule to obtain #dy/(dx)=1/2x^(-1/2)-1/2x^(z3/2)#. Then multiply both sides by dx to get #dy=(1/2x^(-1/2)-1/2x^(z3/2))dx#.

Explanation:

It will be presumed here that you meant exactly what you said, and want the differential #dy# rather than the derivative #dy/(dx)#. If you instead want the derivative, simply ignore the last step.

Recall the power rule states that for any term #ax^b#, with a and b as constants, the derivative is #dy/(dx) = (b)ax^(b-1)#. Recall further that #sqrtx = x^(1/2)# and that #1/sqrtx = 1/x^(1/2) = x^(-1/2)# (because any term #p/x^q = px^(-q)# by the properties of exponents). Thus, if given the function above, we can find its derivative (and differential) as follows:

#y =sqrtx + 1/sqrtx= x^(1/2)+x^(-1/2)#.

#-> dy/(dx) = d/(dx)x^(1/2)+ d/(dx)x^(-1/2) = 1/2x^(-1/2) - 1/2x^(-3/2)#

#dy/(dx)= 1/2x^(-1/2) - 1/2x^(-3/2)#

This, we now have the derivative. If we want the differential #dy#...

#dy/(dx)= 1/2x^(-1/2) - 1/2x^(-3/2) -> dy = ( 1/2x^(-1/2) - 1/2x^(-3/2))dx#