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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

y = x^{2} - 2, (- 3, 7)

3 x y^{^{'}} = y^{2} - y - 2

f (x) = ln (3 x)

\int \frac{1}{( 4 - x ^{2} )} d x

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