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

The derivative of 2x^2+xy-y^2 is 4x+y

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

The first derivative of 2x^2+xy-y^2 is 4x+y

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derivative of

2 x^{2} + x y - y^{2}

4 x + y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 4 x + y - 0

Simplify

= 4 x + y

2 1 - y

\int x 3 81 + x^{2} d x

\frac{\partial}{\partial x} (7 - 2 x^{2} - y^{2})

\int 20 e^{u} d u

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

What is the derivative of 2x^2+xy-y^2 ?
The derivative of 2x^2+xy-y^2 is 4x+y
What is the first derivative of 2x^2+xy-y^2 ?
The first derivative of 2x^2+xy-y^2 is 4x+y