How do you differentiate #g(u)=sqrt2u+sqrt(3u)#?

1 Answer
Jim H
Feb 5, 2017

#g'(u) = sqrt2 + sqrt3/(2sqrtu)#

Explanation:

#sqrt2u# is a constant times #u#, so its derivative (with respect to #u#) is the constant.

If we want to use just the power rule (no chain rule) we'll rewrite

#sqrt(3u) = sqrt3 sqrtu#.

Now we can find #d/(du)(sqrtu)#

#d/(du)(sqrtu) = d/(du)(u^(1/2)) = 1/2 u^(-1/2) = 1/(2u^(1/2)) = 1/(2sqrtu)#.

That is #d/(du) (sqrtu) = 1/(2sqrtu)# (See Note below.)

Therefore,

#d/(du)(sqrt(3u)) = d/(du)(sqrt3 sqrtu)#

# = sqrt3 d/(du)(sqrtu)#

# = sqrt3 1/(2sqrtu)#

# = sqrt3/(2sqrtu)#

For #g(u) = sqrt2u+sqrt(3u)#, we have

#g'(u) = sqrt2 + sqrt3/(2sqrtu)#.

Note

#d/dx(sqrtx) = 1/(2sqrtx)#

I understand that, early on, there seem to be a lot of differentiation rules to memorize, but I suggest eventually memorizing this one also. The square root arises in many problems.
(This is largely because of the Pythagorean Theorem. If we know distances in perpendicular directions, then we use the square root to find the straight line distance.)

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