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

Question

Scarlett Allison · Accepted Answer

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

#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.)

Elijah Abram · Answer

undefined To differentiate g(u) = √(2u) + √(3u), you can use the chain rule:

g'(u) = (1/2)(2u)^(-1/2) * 2 + (1/2)(3u)^(-1/2) * 3

Simplify:

g'(u) = u^(-1/2) + (3/2)u^(-1/2)

Combine like terms:

g'(u) = (1 + 3/2)u^(-1/2)

Simplify further:

g'(u) = (5/2)u^(-1/2)

So, the derivative of g(u) with respect to u is g'(u) = (5/2)u^(-1/2).