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

The derivative of 2x^2+8x is 4x+8

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

The first derivative of 2x^2+8x is 4x+8

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

4 x + 8

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 4 x + 8

x (x^{2} - 11 x + 35), x^{2}

\frac{d}{d x} (\frac{3}{4} x^{12})

y^{^{'}} = - \frac{1}{x} y^{2} - \frac{1}{x} y

x^{\frac{2}{3}} + y^{\frac{2}{3}} = 5, (- 8, - 1)

\int_{1}^{3} x^{4} ln (x) d x

What is the derivative of 2x^2+8x ?
The derivative of 2x^2+8x is 4x+8
What is the first derivative of 2x^2+8x ?
The first derivative of 2x^2+8x is 4x+8