Find dy

[kggirl]

if y = x^2/1+x^2, find dy

my answer is 2x/2x dx which equals to 1dx


is this correct?

 

[jolly]

yepp

 

[kggirl]

THANK YOU

yepp

THANK YOU!

 

[soroban]

Hello, kggirl!

Sheesh . . . neither of you ever heard of the Quotient Rule ??

If $\displaystyle y\:=\:\frac{x^2}{1\,+\,x^2}$, find $\displaystyle dy$.

Differentiate: .\(\displaystyle \L\frac{dy}{dx}\:=\:\frac{(1\,+\,x^2)\cdot2x\,-\,x^2\cdot2x}{(1\,+\,x^2)^2}\:=\:\frac{2x\,+\,2x^3\,-\,2x^3}{(1\,+\,x^2)^2}\:=\:\frac{2x}{(1\,+\,x^2)^2}\)

Therefore: .\(\displaystyle \L dy\;=\;\frac{2x}{(1\,+\,x^2)^2}\,dx\)

 

[kggirl]

thank you!