find the differential of each function y=1+2 u/1+3u

Question

y=(1+2u)/(1+3u)

Answer 1

The differential comes from taking the derivative of both sides without doing it with respect to a specific variable (so you will need to use the chain rule where it applies, but more on this below). Let's begin:

 

 

Derivative of left side = dy

 

Derivative of right side: here we have to use the quotient rule for derivatives, so let f = 1 + 2u, and g = 1 + 3u

Then quotient rule says that d(f/g) = (g*df - f*dg)/(g²)

 

Let's dissect this in pieces and then put it together:

1) df = derivative of f = 2du since the derivative of a constant number (in our case the number 1) is just 0

2) dg = derivative of g = 3du 

 

Note that in (1) and (2) the derivatives got a du. You may have seen in your class something like "the derivative of x is 1," but keep in mind that what really happened there was the "derivative of x WITH RESPECT TO x" or d/dx(x) = 1. Our case deals with differentials, so it's measuring the change of the variable, but not with respect to any variable, hence d(x) = 1*dx because of the chain rule.

 

So let's piece it all together:

 

dy = (2(1+ 3u)du - 3(1 + 2u)du)/(1 + 3u)² = (2(1+ 3u) - 3(1 + 2u))du/(1 + 3u)2 (All I did was factor out the du here)

 

= (2 + 6u - 3 - 6u)du/(1 + 3u)² = -du/(1 + 3u)² 

 

So, dy = -du/(1 + 3u)²

 

And that's it!!

 

Note that if you divide the du over, you have dy/du = -1/(1 + 3u)² which reads as "the derivative of y with respect to u"

 

Pretty neat, right!