# Find the derivative of y= (1+√t)/(1-√t)?

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### 2 Answers

`y = (1+sqrtt)/(1-sqrtt)`

`y = (1+sqrtt)/(1-sqrtt)*(1+sqrtt)/(1+sqrtt)`

`y = (1-sqrtt)^2/(1-(sqrtt)^2)`

`y = (1-sqrtt)^2/(1-t)`

(dy)/dt

`= ((1-t)(2(1+sqrtt)*(1/(2sqrtt)))-(1+sqrtt)^2(-1)}/(1-t)^2`

`= ((1-t)(1/sqrtt+1)+1+2sqrtt+t)/(1-t)^2`

`= (1/sqrtt+1-sqrtt-t+1+2sqrtt+t)/(1-t)^2`

`= (2+sqrtt+1/sqrtt)/(1-t)^2`

`= (sqrtsqrtt+1/sqrtsqrtt)^2/(1-t)^2`

`= (sqrtt+1)^2/(sqrtt(1-t)^2)`

** Therefore** `(dy)/dt = (sqrtt+1)^2/(sqrtt(1-t)^2)`

**Sources:**

Use the division fule for differentation:

If y = f(t)/g(t)

dy/dt = (f'(t)g(t)- f(t)g'(t))/(g(t))^2

Also, multiply the top and bottom by (1 + sqrt(t)) first

y = (1 + 2*sqrt(t) + t)/(1 - t)

so f(t) = (1 + sqrt(t))^2 = 1 + 2*sqrt(t) +t = 1 +2*t^(1/2) + t

g(t) = 1 -t

Gives f'(t) = 2*1/2*t^(-1/2) + 1 = 1/sqrt(t) + 1

g'(t) = -1

So dy/dt = ((1/sqrt(t) + 1)*(1-t) - (1 + sqrt(t))^2*(-1))/(1-t)^2

= ((1/sqrt(t) + 1)*(1-t) + (1 + sqrt(t))^2)/(1-t)^2

= (1/sqrt(t) + 1)/(1-t) + ((1+sqrt(t))/(1-t))^2