If u-tan-1x-y-1-z and x -3t-y-t3-z-3t2- then du-dt-

The correct option is D 3/1+t^2 u=tan^ 1 ( 3t t^31 3t^2 #10; ) put t=tanθ dt=^2θ dθ u=3θ dudθdθdt=dudt 3cos^2θ =dudt ...

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Existence of Limit

If u=tan-1x...

Question

If $u = {tan}^{- 1} (\frac{x - y}{1 - z})$ and $x = 3 t; y = t^{3}; z = 3 t^{2}$ ,
then $\frac{d u}{d t} = ?$

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

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

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\frac{3}{1 + t^{2}}

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\frac{4}{1 + t^{2}}

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Solution

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Similar questions

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

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

t

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

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

\frac{d^{2} y}{d x^{2}}

t = 0

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}}

\frac{d y}{d x}

t = 2

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