What is the derivative of 2x^2+2y-8 ?

The derivative of 2x^2+2y-8 is 4x+2(dy)/(dx)

What is the first derivative of 2x^2+2y-8 ?

The first derivative of 2x^2+2y-8 is 4x+2(dy)/(dx)

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\frac{d}{d x} (2 x^{2} + 2 y - 8)

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

Solution steps

\frac{d}{d x} (2 x^{2} + 2 y - 8)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 x^{2}) + \frac{d}{d x} (2 y) - \frac{d}{d x} (8)

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

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

\frac{d}{d x} (8) = 0

= 4 x + 2 \frac{d y}{d x} - 0

Simplify

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

x \to + 1 + lim (\frac{2}{x} - \frac{1}{ln ( x )})

\frac{x ^{2}}{6 + x}

\frac{\partial}{\partial x} (x^{2} - y^{2} + z^{2} - 2 z - 4)

f (x) = x^{2} - 8, (- 5, 17)

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

What is the derivative of 2x^2+2y-8 ?
The derivative of 2x^2+2y-8 is 4x+2(dy)/(dx)
What is the first derivative of 2x^2+2y-8 ?
The first derivative of 2x^2+2y-8 is 4x+2(dy)/(dx)