Do not understand this (parameter curves)

[MathNoob94]

Hi so my textbook has this explanation:

\(\displaystyle \dfrac{ \dfrac{d}{dt} \left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}\, =\, \dfrac{\dfrac{d}{dt}\, \dfrac{3}{2}\, \left(t\, -\, \dfrac{1}{t}\right)}{2t}\, =\, \dfrac{\dfrac{3}{2}\, \left(1\, +\, \dfrac{1}{t^2}\right)}{2t}\, =\, \dfrac{3\, (t^2\, +\, 1)}{4t^3}\)

However I do not understand the last two steps on how they go from (3/2(1+1/t^2))/(2t) to 3(t^2+1)/(4t^3). Did they take the second derivative or square it??

Thank you

 

[Deleted member 4993]

Hi so my textbook has this explanation:

\(\displaystyle \dfrac{ \dfrac{d}{dt} \left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}\, =\, \dfrac{\dfrac{d}{dt}\, \dfrac{3}{2}\, \left(t\, -\, \dfrac{1}{t}\right)}{2t}\, =\, \dfrac{\dfrac{3}{2}\, \left(1\, +\, \dfrac{1}{t^2}\right)}{2t}\, =\, \dfrac{3\, (t^2\, +\, 1)}{4t^3}\)

However I do not understand the last two steps on how they go from (3/2(1+1/t^2))/(2t) to 3(t^2+1)/(4t^3). Did they take the second derivative or square it??

Those parentheses are important!

can you simplify:

\(\displaystyle \displaystyle{\frac{\frac{3}{2}(A)}{2B} \ = \ ??}\)