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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

y = 4 - x^{2}, (1, 2)

\frac{d y}{d x} = \frac{y}{y + x}

\frac{d}{d x} (2 x^{2} y + 4)

\frac{d}{d x} (\frac{x ^{2} - 3 x - 4}{x - 2})

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