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

The derivative of 2x^2+y^2 is 4x+2y(dy)/(dx)

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

The first derivative of 2x^2+y^2 is 4x+2y(dy)/(dx)

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 x^{2}) = 4 x

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

= 4 x + 2 y \frac{d y}{d x}

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

\frac{d y}{d x} = 3 x y

\int_{0}^{1} 14 cos (x^{2}) d x

\frac{2}{( s + 2 ) ( s + 3 )}

\int_{0}^{2 π} x d x

What is the derivative of 2x^2+y^2 ?
The derivative of 2x^2+y^2 is 4x+2y(dy)/(dx)
What is the first derivative of 2x^2+y^2 ?
The first derivative of 2x^2+y^2 is 4x+2y(dy)/(dx)