Question

x=sin^3t y= cos^3t then dy/dx​

Answer 1

Given ,

The two functions are

$\tt x = {sin}^{3} (t) \: \: and \: \: y = {cos}^{3} (t)$

Differentiating x wrt t , we get

$\tt \implies \frac{dx}{dt} = \frac{d \{ {sin}^{3} (t) \}}{dt}$

$\tt \implies\frac{dx}{dt} = 3 {sin}^{2} (t) \frac{d \{sin(t) \}}{dt}$

$\tt \implies\frac{dx}{dt} = 3 {sin}^{2} (t) cos(t)$

Similarly , Differentiating y wrt t , we get

$\tt \implies \frac{dy}{dt} = \frac{d \{ {cos}^{3} (t) \}}{dt}$

$\tt \implies \frac{dy}{dt} = 3 {cos}^{2} (t) \frac{d \{cos(t) \}}{dt}$

$\tt \implies\frac{dy}{dt} = - 3 {cos}^{2} (t) sin(t)$

Now ,

$\boxed{ \tt{\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } }}$

Thus ,

$\tt \implies \frac{dy}{dx} = - \frac{3 {cos}^{2} (t)sin(t)}{3 {sin}^{2}(t)cos(t) }$

$\tt \implies\frac{dy}{dx} = - \frac{cos(t)}{sin(t)}$

$\tt \implies\frac{dy}{dx} = - cot(t)$

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Answer 2

Given ,

The two functions are

$\tt x = {sin}^{3} (t) \: \: and \: \: y = {cos}^{3}$

Differentiating x wrt t , we get

$\tt \implies \frac{dx}{dt} = \frac{d \{ {sin}^{3} (t) \}}{dt}</p><p> \\ </p><p>\tt \implies\frac{dx}{dt} = 3 {sin}^{2} (t) \frac{d \{sin(t) \}}{dt} </p><p> \\ </p><p>\tt \implies\frac{dx}{dt} = 3 {sin}^{2} (t) cos(t)$

Similarly , Differentiating y wrt t , we get

$\tt \implies \frac{dy}{dt} = \frac{d \{ {cos}^{3} (t) \}}{dt}</p><p></p><p> \\ </p><p>\tt \implies \frac{dy}{dt} = 3 {cos}^{2} (t) \frac{d \{cos(t) \}}{dt}</p><p> \\ </p><p> </p><p></p><p>\tt \implies\frac{dy}{dt} = - 3 {cos}^{2} (t) sin(t)$

Now ,

$\boxed{ \tt{\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } }} </p><p>$

Thus ,

$\tt \implies \frac{dy}{dx} = - \frac{3 {cos}^{2} (t)sin(t)}{3 {sin}^{2}(t)cos(t) }</p><p> </p><p> \\ </p><p></p><p>\tt \implies\frac{dy}{dx} = - \frac{cos(t)}{sin(t)}</p><p> </p><p> \\ </p><p></p><p>\tt \implies\frac{dy}{dx} = - cot(t)$

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