A maths problem in Differentiation......Plz Help?

Question

If
A=y + 1/(y+3) find
dA/dy, simplyfying the answer

Answer 1

Use the quotient rule!

A = (y+1)/(y+3)

[[Quotient rule: dy/dx = (u'v - uv') / v^2

u and v are the numerator and denominator of the function, respectively]]

u= y+1
u'=1

v= y+3
v'=1

dA/dy = [ (y+3) - (y +1) ] / [ (y+3)^2]
dA/dy = 2 / [ (y+3)^2 ]
dA/dy = 2 / (y^2+ 6y + 9)

That's your answer! Hope that helped!

Answer 2

You have two additive terms in y = A, so you differentiate term by term. The second term in a bit tricky, since it is, expressed as a numerator,
(y+3)-1. However, the power rule applies, and you have a "1" coefficient for y", so you get
dA/dy = 1 (for d(y)/dy) -1 (y+3)^-2

Answer 3

dA/dy=1-1/((y+3)^2)=((y+4)*(y+2))/((y+3)^2)