What is the derivative of 2-y^2+x^2 ?

The derivative of 2-y^2+x^2 is 2x-2y(dy)/(dx)

What is the first derivative of 2-y^2+x^2 ?

The first derivative of 2-y^2+x^2 is 2x-2y(dy)/(dx)

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2) = 0

\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}

\frac{d}{d x} (x^{2}) = 2 x

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

Simplify

= 2 x - 2 y \frac{d y}{d x}

(1, 1), (- 4, - 3)

y^{^{'}} = - 4 x y^{3}

\int \frac{x - 3}{x ^{3} + 2 x ^{2}} d x

\frac{d}{d x} (5 x^{2} - 3 x + 2)

\frac{2}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

What is the derivative of 2-y^2+x^2 ?
The derivative of 2-y^2+x^2 is 2x-2y(dy)/(dx)
What is the first derivative of 2-y^2+x^2 ?
The first derivative of 2-y^2+x^2 is 2x-2y(dy)/(dx)