If u-cos-1x-y-x-4t3-y-3t- then du-dt-

The correct option is B 3/√(1 t^2) put #160;t=sinθ dt=cosθ #160; dθ u=cos^ 1( sin(+3θ )) u=π2 3θ dudt=dudθdθdt = 3(θ ...

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Parametric Differentiation

If u=cos-1x...

Question

If $u = {cos}^{- 1} (x - y); x = 4 t^{3}; y = 3 t$ , then $\frac{d u}{d t} =$

- \frac{1}{\sqrt{1 - t^{2}}}

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- \frac{2}{\sqrt{1 - t^{2}}}

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- \frac{3}{\sqrt{1 - t^{2}}}

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- \frac{4}{\sqrt{1 - t^{2}}}

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Solution

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Similar questions

x = {sin}^{- 1} (3 t - {4 t}^{3})

y = {cos}^{- 1} (\sqrt{1 - t^{2}})

\frac{d y}{d x}

x = \frac{1 - t^{2}}{1 + t^{2}}, y = \frac{2 t}{1 + t^{2}}

\frac{d y}{d x} = - \frac{x}{y}

x

\frac{t^{2}}{1 + t^{2}}

y

\frac{2 a t}{1 + t^{2}}

\frac{d y}{d x}

x = \frac{2 t}{1 + t^{2}}

y = \frac{1 - t^{2}}{1 + t^{2}}

t

x = \frac{2 t}{1 + t^{2}}

y = \frac{1 - t^{2}}{1 + t^{2}}

\frac{d y}{d x}

t = 2

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