What is the derivative of 2x^2+4y^2 ?

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

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

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

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derivative of

2 x^{2} + 4 y^{2}

4 x

Solution steps

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

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

\frac{d}{d x} (4 y^{2}) = 0

= 4 x + 0

Simplify

= 4 x

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

\int 15.7 - 0.2 x d x

\int_{4}^{9} 3 x d x

\int 3 x^{3} cos (x) d x

t an g e n t f (x) = x^{2} - 5 x + 5, at x = 0

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