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

x \to 1 + lim (ln (x^{x - 1}))

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

\frac{d}{d x} (108 x)

x \to 1 lim (\frac{x}{x + 1})

f (x) = 0.8 e^{x} arctan (x)

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