Finding derivative of y + 1/(y+3)

[brandon26]

Hi, I am stuck on this question. Can someone please help me as quick as possible, I am revising for an exam on Monday.

If A= y + 1/(y+3) , find dA/dy.

I have been tryin to solve this question but I keep getting the wrong answer.

 

[nrm]

What have you tried so far?

 

[TD]

I assume you know that differentiation is lineair? So (f+g)' = f' + g'.
You should at least be able to do a part then

 

[brandon26]

Well I tried to make the equation simpler first. y + 1/(y+3) = (y^2 + 3y + 1)(y+3)^-1 right so far?

 

[brandon26]

I assume you know that differentiation is lineair? So (f+g)' = f' + g'.
You should at least be able to do a part then

Sorry, confused. I just don't know how to start the question in the first place.

 

[TD]

Well I tried to make the equation simpler first. y + 1/(y+3) = (y^2 + 3y + 1)(y+3)^-1 right so far?

That's right but it doesn't make it simpler!
You're probably used to factoring a lot in order to solve equations etc, but when you're going to differentiate (or integrate, later) you'll see that it's easier to differentiate sums than products!

Now, if you need to find the derivative of a sum, you can just take the derivative of each term. In your case (a ' denotes a derivative):

\left( {y + \frac{1}{{y + 3}}} \right)^\prime = \left( y \right)^\prime + \left( {\frac{1}{{y + 3}}} \right)^\prime

 

[brandon26]

That's right but it doesn't make it simpler!
You're probably used to factoring a lot in order to solve equations etc, but when you're going to differentiate (or integrate, later) you'll see that it's easier to differentiate sums than products!
Now, if you need to find the derivative of a sum, you can just take the derivative of each term. In your case (a ' denotes a derivative):
\left( {y + \frac{1}{{y + 3}}} \right)^\prime = \left( y \right)^\prime + \left( {\frac{1}{{y + 3}}} \right)^\prime

That gives me dA/dy= 1 - 1/(y+3)^2. Is that right?

 

[TD]

That gives me dA/dy= 1 - 1/(y+3)^2. Is that right?

That is correct