If u-cosx2-y2 then x- u- x-y- u- y-

The correct option is A 2 cos^ 1u√(1 u^2) cos^ 1u=x^2+y^2 #10; #10; 1√(1 u^2)dudx=2x #10; #10; 1√(1 u^2)dudy=2y #10; #10; ...

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Circular Measurement of Angle

If u=cosx2+...

Question

If $u = cos (x^{2} + y^{2})$ then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} =$

- 2 {cos}^{- 1} u \sqrt{1 - u^{2}}

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\frac{- 2 {cos}^{- 1} u}{\sqrt{1 - 4} u}

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2 u

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2 {cos}^{- 1} u

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Solution

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Similar questions

y = tan (\frac{1}{2} {cos}^{- 1} \frac{1 - u^{2}}{1 + u^{2}} + \frac{1}{2} {sin}^{- 1} \frac{2 u}{1 + u^{2}})

x = \frac{2 u}{1 - u^{2}}

\frac{d y}{d x}

y = t a n (\frac{1}{2} c o s^{- 1} \frac{1 - u^{2}}{1 + u^{2}} + \frac{1}{2} s i n^{- 1} \frac{2 u}{1 + u^{2}})

x = \frac{2 u}{1 - u^{2}}

\frac{d y}{d x}

I = \int t a n^{- 1} \sqrt{(\sqrt{x} - 1)} d x = (u^{2} + 1)^{2} t a n^{- 1} u - \frac{A}{1863} u^{3} - u + C

u = \sqrt{\sqrt{x} - 1}

y = {tan}^{- 1} \frac{u}{\sqrt{1 - u^{2}}}

x = {sec}^{- 1} \frac{1}{2 u^{2} - 1}

u \in (0, \frac{1}{\sqrt{2}})

\cup (\frac{1}{\sqrt{2}}, 1)

2 \frac{d y}{d x} + 1 = 0

u = {tan}^{- 1} (\frac{x^{3} + y^{3}}{x - y})

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?

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