help with derivatives?

Question

Find dz/dy if z= (e^y + 1)/ (y+1)

Answer 1

(u/v)' = (u'v - uv')/v²
d[(e^y + 1)/ (y+1)]/dy = [e^y * (y+1) - (e^y+1)*1] / (y+1)²
= (y*e^y + e^y - e^y - 1) / (y+1)²
= (y*e^y-1) / (y+1)²

Answer 2

z= (e^y + 1)/ (y+1)

dz/dy = d/dy{(e^y + 1)/ (y+1)}
= d/dy(e^y/y+1) + d/dy{1/(y+1))
= d/dy((y+1) * d/dye^y - e^y* d/dy(y+1))/(y+1)^2 + -1/(y-1)^2
=((y+1)e^y-e^y)/(y+1)^2 - 1/(y-1)^2
= ye^y-1/(y+1)^2

Answer 3

I should get best answer since every one else is WRONG!!!!

using quotient rule: f'g - fg' / (g^2)

where f = (e^y + 1)
and g = (y+1)

f' = e^y
g' = 1
g^2 = (y+1)^2

now just put the right things in the right place:

dz/dy = [ (e^y) * (y+1) - (e^y + 1) * (1) ] / [ (y+1)^2 ]

simplifies to:

e^y(y+1 - 1) / (y+1)^2
e^y(y) / (y+1)^2

Answer 4

Use the product and chain rules.

dz/dy = d(e^y + 1)/dy*[1/(y +1)] + d(1/(y+1))/dy*[e^y +1] (Product rule)

= d(e^y + 1)/dy*[1/(y +1)] - d(y+1))/dy*[(e^y +1)/(y + 1)^2] (Chain Rule)

= e^y -(1/(y+1)^2)*1 = e^y/(y +1) - (e^y +1)/(y +1)^2

Answer 5

Use the quotient rule:-

If y = u/v
Then
dy/dx = [v du/dx - u dv/dx] / v^2

So in your case u = e^y + 1 & v = (y+1)
du/dy = e^y & dv/dy = 1

dz/dy = [(y + 1)e^y - (e^y +1)1] / (y + 1)^2

dz/dy = [ ye^y + e^y - e^y - 1] / ( y + 1)^2

dz/dy = [ ye^y - 1] / (y + 1)^2

Answer 6

z= (e^y + 1)/ (y+1)
z' = [e^y(y+1) - (e^y + 1)] / (y+1)^2
z' = [ye^y+e^y - e^y - 1] / (y+1)^2
z' = [ye^y - 1] / (y+1)^2