If z-tanxy-x2-y2 then x- z- x-y- z- y-

The correct option is B 2 #160;tan^ 1z.(1+z^2) #10; tan^ 1z=xy+x^2+y^2 It is of degree n=2 11+z^2 ( ∑xdzdx )=n #160;tan^ 1 z ( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Parametric Differentiation

If z=tanxy+...

Question

If $z = tan (x y + x^{2} + y^{2})$

then $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = ?$

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 {tan}^{- 1} z . (1 + z^{2})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 {tan}^{- 1} z}{1 + z^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

Δ A B C, ∠ A = {sin}^{- 1} (x), ∠ B = {sin}^{- 1} (y) a n d ∠ C = {sin}^{- 1} (z)

x \sqrt{1 - y^{2}} \sqrt{1 - z^{2}} + y \sqrt{1 - x^{2}} \sqrt{1 - z^{2}} + z \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}

{sin}^{- 1} x + {sin}^{- 1} y + {sin}^{- 1} z = π

x \sqrt{1 - x^{2}} + y \sqrt{1 - y^{2}} + z \sqrt{1 - z^{2}} = 2 x y z

x + y + z = x y z

\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \cdot \frac{2 y}{1 - y^{2}} \cdot \frac{2 z}{1 - z^{2}}

x + y + z = 0

\frac{1}{x^{2} + y^{2} - z^{2}} + \frac{1}{y^{2} + z^{2} - x^{2}} + \frac{1}{z^{2} + x^{2} - y^{2}}

s i n^{- 1} x + s i n^{- 1} y + s i n^{- 1} z = π

x \sqrt{1 - x^{2}} + y \sqrt{1 - y^{2}} + z \sqrt{1 - z^{2}} = x y z

Join BYJU'S Learning Program