derivative of y(x) = (t / (t+1))^2

[Becky4paws]

y(x)=(t/t+1)^2

y(x) = t^2/(t+1)^2

y'(x)=(t+1)^2 (2) (2t) - (t^2) (2) (t+1)/(t+1)^4

Y'(x)=8t[t^2(t+1)/(t+1)^2]

I don't think I'll ever understand this...

 

[soroban]

Re: derivatives

Hello, Becky!

You were doing great . . .

\(\displaystyle y(x)\:=\:\left(\frac{t}{t\,+\,1\right)^2\)

\(\displaystyle y(x) \:= \:\frac{t^2}{(t\,+\,1)^2}\)

\(\displaystyle y'(x)\:=\:\frac{(t\,+\,1)^2(2t)\,-\,(t^2)(2)(t\,+\,1)}{(t\,+\,1)^4}\)

We have: \(\displaystyle \,y'(x)\;= \;\frac{2t(t\,+\,1)^2\,-\,2t^2(t\,+\,1)}{(t\,+\,1)^4}\)

Factor: \(\displaystyle \,y'(x)\;= \;\frac{2t(t\,+\,1)\,\left[(t\,+\,1)\,-\,t\right]}{(t\,+\,1)^4} \;= \;\frac{2t(t\,+\,1)(1)}{(t\,+\,1)^4}\;=\;\frac{2t}{(t\,+\,1)^3}\)

 

[stapel]

y(x)=(t/t+1)^2

If y is a function of x ("y(x)"), then wouldn't the derivative with respect to x, dy/dx, equal 0...?

Eliz.