What is (\partial)/(\partial x)(y^2+x^2(1-xy)^{-2}) ?

The answer to (\partial)/(\partial x)(y^2+x^2(1-xy)^{-2}) is (2x)/((-yx+1)^3)

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(\partial)/(\partial x)(y^2+x^2(1-xy)^{-2})

\frac{\partial}{\partial x} (y^{2} + x^{2} (1 - x y)^{- 2})

\frac{2 x}{( - y x + 1 ) ^{3}}

Solution steps

\frac{\partial}{\partial x} (y^{2} + x^{2} (1 - x y)^{- 2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (x^{2} (1 - x y)^{- 2})

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (x^{2} (1 - x y)^{- 2}) = \frac{2 x}{( - y x + 1 ) ^{3}}

= 0 + \frac{2 x}{( - y x + 1 ) ^{3}}

Simplify

= \frac{2 x}{( - y x + 1 ) ^{3}}

\frac{\partial}{\partial x} (y^{2} - y)

f (r) = π r^{2}

\frac{d}{d x} (e^{c x})

\int_{- 1}^{3} (x^{2} - 3) d x

2 cos (2 x) + 2 sin (2 x) + 4 x

What is (\partial)/(\partial x)(y^2+x^2(1-xy)^{-2}) ?
The answer to (\partial)/(\partial x)(y^2+x^2(1-xy)^{-2}) is (2x)/((-yx+1)^3)