differentiation question.. (1 Viewer)

[skillz]

Can someone please help me with this.

(3x+1)^(3/2)*(2x+4)
where, U= (3x+1)^3/2 and V = 2x+4

thank you.

 

[darkliight]

This one will require us to use the chain rule and the product rule. U will be the one that needs the chain rule as it's a function of a function (are you ok with that?).

We'll do U first.

Set
s = 3x + 1
then U = s^(3/2)

ds/dx = 3
dU/ds = 3/2 * s^(1/2) = 3/2 * (3x+1)^(1/2)

dU/dx = dU/ds * ds/dx (by the chain rule)
= 3 * 3/2 * (3x+1)^(1/2)
= 9 * (3x+1)^(1/2)/2

--------------------------------------------------------------

Now we'll do V.

dV/dx = 2

--------------------------------------------------------------

Now that we have dU/dx and dV/dx we can use our product rule.

d/dx (3x+1)^(3/2) * (2x+4) = d/dx U * V = dU/dx * V + U * dV/dx (by the product rule)
= 9 * (3x+1)^(1/2) * (2x+4)/2 + 2 * (3x+1)^(3/2)

A complex one, but I hope that helps.

Cheers.

 

[abcd9146]

i think your right so far, but you havent finished the quection yet.
you have:
= 9 * (3x+1)^1/2 * (2x+4)/2 + 2 * (3x+1)^3/2

take out (3x+1)^1/2=

= (3x+1)^1/2[(9/2)(2x+4) + 2(3x+1)]
= (3x+1)^1/2(9x+18+6x+2)
= (3x+1)^1/2(15x+20)
= 5(3x+1)^1/2(3x+4)

 

[PLooB]

i think your right so far, but you havent finished the quection yet.
you have:
= 9 * (3x+1)^1/2 * (2x+4)/2 + 2 * (3x+1)^3/2

take out (3x+1)^1/2=

= (3x+1)^1/2[(9/2)(2x+4) + 2(3x+1)]
= (3x+1)^1/2(9x+18+6x+2)
= (3x+1)^1/2(15x+20)
= 5(3x+1)^1/2(3x+4)

Correct.

Quickest way is to use the product rule.