What is the derivative of 2y^2+3x ?

The derivative of 2y^2+3x is 4y(dy)/(dx)+3

What is the first derivative of 2y^2+3x ?

The first derivative of 2y^2+3x is 4y(dy)/(dx)+3

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

4 y \frac{d y}{d x} + 3

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (3 x) = 3

= 4 y \frac{d y}{d x} + 3

\frac{\partial}{\partial y} (x^{3} - 6 x y + 8 y^{3})

\frac{\partial}{\partial x} (cos (7 x + 4 y + z))

x \to 0 lim (\frac{ln ( 1 - 3 x )}{x})

\int 15 x^{2} - cos (x) + C x + C d x

f (x) = \frac{( x ^{2} - 4 x )}{x + 1}

What is the derivative of 2y^2+3x ?
The derivative of 2y^2+3x is 4y(dy)/(dx)+3
What is the first derivative of 2y^2+3x ?
The first derivative of 2y^2+3x is 4y(dy)/(dx)+3