Difficult Calculus Question (Differentiation) I stumbled across this question in my maths textbook and have no idea how to solve it. I have looked at the answers and don't understand how to get it from the question. It's in the chapter titled 'The Quotient Rule' but am not sure how to apply it to this question:
Find the normal to the curve $x = \frac{t}{t+1}$ and $y = \frac{t}{t-1}$ at the point T where t = 2.
I tried converting this into one equation ($y = \frac{x(t+1)}{t-1}$) but don't see how to get the answer which says:
$$\frac{dy}{dx} = \frac{-(t+1)^2}{(t-1)^2}$$
and that $ T = (\frac23,2)$, $3x - 27y + 52 = 0$
I'd appreciate any pointers.
 A: Recall that the chain rule gives $\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}$. So as long as $\frac{dx}{dt} \neq 0$, we can solve for $\frac{dy}{dx}$, yielding the equation:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ 
This should give you the slope of the tangent line to the parametric curve you specified.
A: We have: $\dfrac{t+1}{t} = \dfrac{1}{x} \Rightarrow 1+\dfrac{1}{t} = \dfrac{1}{x}$. Similarly, $\dfrac{t-1}{t} = \dfrac{1}{y} \Rightarrow 1-\dfrac{1}{t} = \dfrac{1}{y}$. Add the $2$ equations to get: $2=\dfrac{1}{x}+\dfrac{1}{y} \Rightarrow y = \dfrac{x}{2x-1}$. From this you can directly find $\dfrac{dy}{dx}$,and to completely answer the question.
A: Hint:
$x = \frac{t}{t+1}$ and $y = \frac{t}{t-1}$
$$\frac{dx}{dt}=\frac{t+1-t}{(t+1)^2}=\frac{1}{(t+1)^2}$$
$$\frac{dy}{dt}=\frac{t-1-t}{(t-1)^2}=\frac{-1}{(t-1)^2}$$
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-(t+1)^2}{(t-1)^2}$$
A: Sorry if this working isn't top notch, but I'm only a 16 year old student! It's a bit late too as this is the first time I've used LaTeX!
By, the quotient rule, we have:
$$\frac{\text{d}x}{\text{d}t}=\frac{1(t+1)-t(1)}{(t+1)^{2}}=\frac{-1}{(t-1)^{2}}$$
$$\frac{\text{d}y}{\text{d}t}=\frac{1(t-1)-t(1)}{(t-1)^{2}}=\frac{-1}{(t-1)^{2}}$$
By the chain rule:
$$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}y}{\text{d}t}\cdot\frac{\text{d}t}{\text{d}x}$$
$$\therefore\frac{\text{d}y}{\text{d}x}=\frac{-1}{(t-1)^{2}}\cdot\frac{(t+1)^{2}}{1}$$
$$\Rightarrow\frac{\text{d}y}{\text{d}x}=\frac{-(t+1)^{2}}{(t-1)^{2}}$$
A: X= t/(t+1) 
Y= t/(t-1)
dx/dt = (t+1)-t / (t+1)²
dx/dt= 1/(t+1)²
dy/dt = (t-1)-t/(t-1)²
dy/dt= -1/(t-1)²
dy/dx = dy/dt × dt/dx
note, dt/dx is d inverse of dx/dt
So, 
dy/dx = -1/(t-1)² × (t+1)²
dy/dx = -(t+1)²/(t-1)²
When t= 2,
dy/dx = -(2+1)²/(2-1)²
dy/dx = -9
Note, dy/dx = gradient = m
So,
m = -9
Gradient of normal
m'= -1/m
So, 
m' = -1/-9
m'= 1/9
Equation of normal =
Y-y = (X-x)m'
Recall,  x= t/(t+1)
y= t /(t-1)
Since t =2
Substitute for t
x= 2/2+1
x=2/3
y=2/2-1
y=2
Back to the equation of normal
Substitute for x and y and m'
Y-2 = (X-⅔)×1/9
Y-2=X/9 - 2/27
Y-2=(3X-2)/27
27(Y-2) = 3X-2
27Y-54=3X-2
27Y=3X+52
SOLVED
A: This seems to be a question of Parametric derivative.
$$(x,y)=\left(\frac{t}{t+1},\frac{t}{t-1}\right)$$
$$\frac{dx}{dt}=\frac{d(t(t+1)^{-1})}{dt}=t(-1)(t+1)^{-2}+(t+1)^{-1}=\frac{-t}{(t+1)^2}+\frac{1}{(t+1)^2}$$
Similarly,
$$\frac{dy}{dt}=\frac{d(t(t-1)^{-1})}{dt}=\frac{-t}{(t-1)^2}+\frac{1}{t-1}$$
Therefore,$$\frac{dy}{dx}=\frac{\frac{-t}{(t-1)^2}+\frac{1}{t-1}}{\frac{-t}{(t+1)^2}+\frac{1}{t+1}}=\frac{-t+t-1}{(t-1)^2}\times \frac{(t+1)^2}{-t+t+1}=-\left(\frac{t+1}{t-1}\right)^2$$
This is the slope of the tangent.
The slope of the normal is $\left(\frac{t-1}{t+1}\right)^2$. Since $t=2$, therefore the slope of normal is $\frac{1}{9}$.
Coordinates of point $T=\left(\frac{2}{2+1},\frac{2}{2-1}\right)=\left(\frac{2}{3},2\right)$
So, the equation of normal is $y=\frac{1}{9}x+c$. Since normal passes through the point T, therefore,
$$2=\frac{1}{9}\times \frac{2}{3}+c \implies c=\frac{52}{27}$$
Therefore the equation of normal is $y=\frac{1}{9}x+\frac{52}{27}\implies 3x-27y+52=0$.
