Question

X=t square, y=t cube then d square y upondx square

Answer 1

it's (B)


please mark me as brainlist

Answer 2

hey dear

here your answer

$x = {t}^{2} \\ y = {t}^{3} \\ \\ x = {t}^{2} \: \: \: \: \therefore \: \: t \: = \sqrt{x} = {x}^{ \frac{1}{2} } \\ \\ y = {t}^{3} = {( {x}^{ \frac{1}{2} }) }^{3} = {x}^{ \frac{3}{2} } \\ \\ \frac{dy}{dx} = \frac{d( {x}^{ \frac{3}{2} }) }{dx} \: \: \: \: \: \because \: y \: = {x}^{ \frac{3}{2} } \\ \\ \: \: \: \: \: \: \: = \frac{3}{2} {x}^{ \frac{3}{2} - 1 } = \frac{3}{2} {x}^{ \frac{1}{2} } \\ \\ \frac{ {d}^{2}y}{d {x}^{2} } = \frac{3}{2} \frac{d( {x}^{ \frac{1}{2} }) }{dx} \\ \\ \frac{ {d}^{2}y }{d {x}^{2} } = \frac{3}{2} \times \frac{1}{2} {x}^{ \frac{1}{2} - 1} = \frac{3}{4} {x}^{ \frac{ - 1}{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: = \frac{3}{4} \times \frac{1}{ \sqrt{x} } \\ \\ \: \: \: \: \: \: \: \: \: \: = \frac{3}{4} \times \frac{1}{ \sqrt{ {t}^{2} } } \\ \\ \: \: \: \: \: \: \: \: \: \: = \frac{3}{4t} ...........your \: answer \: \\ \\ hope \: it \: helps \: you$