Differentiate the parametric function and find $dy/dx$ and $d^2y/dx^2$

Question

Differentiate the parametric function and find $\frac{\mathrm dy}{\mathrm dx}$ and $\frac{\mathrm d^2y}{\mathrm dx^2}$ in terms of "$t$" when:

$ x = \frac{1}{t-1}$ and $y = \frac{1}{t+1}$

I have first started by finding $\frac{\mathrm dy}{\mathrm dx}$ by finding $\frac{\mathrm dx}{\mathrm dt}$ and $\frac{\mathrm dy}{\mathrm dt}$ which comes to

$$\frac{\mathrm dx}{\mathrm dt}=\ln |t-1|$$ and $$\frac{\mathrm dy}{\mathrm dt}=\ln |t+1|$$

and used $\frac{\mathrm dy}{\mathrm dx} =\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt} $ ,which gives

$$\frac{\mathrm dy}{\mathrm dx}=\frac{\ln|t+1|}{\ln|t-1|}$$ now if I divide them by each other doesn't it equal to $0$? What have I done wrong here?

Answer 1

Given $$\displaystyle x = \frac{1}{t-1}.$$ So $\displaystyle \frac{dx}{dt} = -\frac{1}{(t-1)}$ and $$\displaystyle y = \frac{1}{t+1}$$ So $\displaystyle \frac{dx}{dt} = -\frac{1}{(t+1)}$

Now $$\displaystyle \frac{1}{y}-\frac{1}{x} = 2\;,$$ Now Diff. both side w. r to $x$

$$\displaystyle -\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{x^2} = 0\Rightarrow \frac{dy}{dx} = \frac{y^2}{x^2}$$

Now Again Diff. w. r to $x\;,$ We get

$$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \left[\frac{y^2}{x^2}\right] = \frac{x^2\cdot 2y\frac{dy}{dx}-y^2 \cdot 2x}{x^4} = \frac{2y^3-2xy^2}{x^4}$$

Above we have put value f $\displaystyle \frac{dy}{dx} = \frac{y^2}{x^2}$ So we get $$\displaystyle \frac{d^2y}{dx^2} = \frac{2y^3-2xy^2}{x^4}$$

Answer 2

Given $\displaystyle x = \frac{1}{t-1}$ and $\displaystyle y = \frac{1}{t+1}$

the differentiation of $x^n$ is $nx^{n-1}$

here $ x= \frac{1}{t-1} = (t-1)^{-1}$

So $ \frac{dx}{dt}= \frac{d(t-1)^{-1}}{dt}$

$= \frac{d(t-1)^{-1}}{d(t-1)} . \frac{d(t-1)}{dt}$ $=(-1)(t-1)^{-1-1}.(1)$ $=-(t-1)^{-2}$ $=\frac{-1}{(t-1)^{2}}$

Similarly you can do with Y and then for $dx^{2}$ and $dy^{2}$