Question

find dy/dx for y=t^2 and x=2t^3

Answer 1

Hi this is parametric differentiation.
$y = {t}^{2} \\ \frac{dy}{dt} = 2t \\ similarly \\ x = 2 {t}^{3} \\ \frac{dx}{dt} = 6 {t}^{2} \\now \: \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \\ = \frac{2t}{6 {t}^{2} } \\ = \frac{1}{3t}$
now given x =2t^3.
Therefore t = cubic root of (x/2).
Therefore dy/dx=1/(3×cubic root of (x/2))

Hope this helps.

Answer 2

in the above pic last step is wrong ... the ryt one is 2t/ 6t^2
=== Dy/dx = 1/3t

due to my phns's problem this happens so I take ss nd post a pic ...

I hope u get it... then pls mark it as brainliest ..