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

The derivative of 8x^2+x is 16x+1

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

The first derivative of 8x^2+x is 16x+1

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

16 x + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (8 x^{2}) = 16 x

\frac{d x}{d x} = 1

= 16 x + 1

n = 1 \sum \infty (\frac{1}{2})^{n - 1}

\int \frac{t ^{2}}{t ^{4} + 1} d t

\frac{\partial}{\partial y} (6 e^{yz} cos (x yz))

\frac{d}{d x} (1 1 + x)

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

What is the derivative of 8x^2+x ?
The derivative of 8x^2+x is 16x+1
What is the first derivative of 8x^2+x ?
The first derivative of 8x^2+x is 16x+1