Find the value of dy/dx , if x= 2 cos square t, y= 2 sin square t

1.	Find the value of dy/dx , if x= 2 cos square t, y= 2 sin square t
Answer» ANSWER :– $\\ \longrightarrow \: \: \large { \red{ \boxed{ \bold{ \dfrac{dy}{dx} = - <klux>1</klux> }}}} \\$ EXPLANATION :– GIVEN :– • ${ \bold{ x = <klux>2</klux> \: {cos}^{2}t }}$ and ${ \bold{ y = 2 \: {sin}^{2}t }} \\$ TO FIND :– • The value of ${ \bold{ \: \dfrac{dy}{dx} = ? }} \\$ SOLUTION :– • We know that – $\\ \implies { \pink{ \boxed{ \bold{ \dfrac{dy}{dx} = \dfrac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} } }}}} \\$ • So , let's find ${ \bold{ \: \dfrac{dy}{dt} \: \: - }} \\$ $\\ { \bold{ \implies y = 2 \: {sin}^{2}t }}$ • Now DIFFERENTIATE with respect to 't' – $\\ { \bold{ \implies \frac{dy}{dt} = 2 \times \: 2{sin}t \times cost }}$ $\\ { \bold{ \implies \frac{dy}{dx} = 4 \: {sin}t . cost \: \: \: \: - - - -eq.(1) }}$ $\\ { \bold{ \implies x = 2 \: {cos}^{2}t }}$ • Now Differentiate with respect to 't' – $\\ { \bold{ \implies \frac{dx}{dt} =2 \times 2 \: {cos}t \times ( - sint ) }}$ $\\ { \bold{ \implies \frac{dx}{dt} =4\: {cos}t \times ( - sint ) \: \: \: - - - - eq.(2) }}$ • Now divided eq.(1) and eq.(2) – $\\ \because \: \: \: { \bold{ \dfrac{dy}{dx} = \dfrac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} } }} \\$ • Put the values from eq.(1) and eq.(2) – $\\ \implies \: \: \: { \bold{ \dfrac{dy}{dx} = \dfrac{ 4 sin(t) cos(t) }{4 cos(t) [ - sin(t) ] } }} \\$ $\\ \implies \: \: \large { \red{ \boxed{ \bold{ \dfrac{dy}{dx} = - 1 }}}} \\$ $\\ \: \: \large { \bold{ \underline{Used \: \: formula} : - }} \\$ $\\ \: \: { \bold{(1) \: \: \frac{d( {x}^{n} )}{dx} = n {x}^{n - 1} }} \\$ $\\ \: \: { \bold{(2) \: \: \frac{d[sin( \theta)]}{dx} = cos( \theta) }} \\$ $\\ \: \: { \bold{(3) \: \: \frac{d[cos( \theta)]}{dx} = - \: sin( \theta) }} \\$

Answer»

ANSWER :–

$\\ \longrightarrow \: \: \large { \red{ \boxed{ \bold{ \dfrac{dy}{dx} = - <klux>1</klux> }}}} \\$