Question

Find the first derivative for y=(x+1)^2 / 1+x^2

Answer 1

quotient rule....

Answer 2

with a chain rule

Answer 3

I odnt get it =////

Answer 4

dont*

Answer 5

do you know the quotient rule?

Answer 6

u= (x+1)^2

du/dx = 2(x+1)

v= 1+x^2
dv/dx = 2x

dy/dx = (inside-outside) / (bottom squared )

dy/dx = [ 2(x+1)(1+x^2) -2x(x+1)^2] / ( 1+x^2)^2

dy/dx = (x+1) [ 2(1+x^2) -2x(x+1) ] / (1+x^2)^2

Answer 7

= (x+1)( 2 - 2x) / (1+x^2)^2

Answer 8

\[= \frac{ 2(x+1)(1-x) }{(1+x^2)^2}\]

Answer 9

\[= \frac{-2(x^2-1)}{(1+x^2)^2}\]

Answer 10

Thanks mate