What is the derivative of g(x)=1+6x+3x^2 ?

The derivative of g(x)=1+6x+3x^2 is 6x+6

What is the first derivative of g(x)=1+6x+3x^2 ?

The first derivative of g(x)=1+6x+3x^2 is 6x+6

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derivative of

g (x) = 1 + 6 x + 3 x^{2}

6 x + 6

Solution steps

\frac{d}{d x} (1 + 6 x + 3 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1) + \frac{d}{d x} (6 x) + \frac{d}{d x} (3 x^{2})

\frac{d}{d x} (1) = 0

\frac{d}{d x} (6 x) = 6

\frac{d}{d x} (3 x^{2}) = 6 x

= 0 + 6 + 6 x

Simplify

= 6 x + 6

\int \frac{x}{x ^{2} + 4 x + 20} d x

\frac{\partial}{\partial y} (cos (x) cos (y))

\int e^{4 x + 8} d x

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

What is the derivative of g(x)=1+6x+3x^2 ?
The derivative of g(x)=1+6x+3x^2 is 6x+6
What is the first derivative of g(x)=1+6x+3x^2 ?
The first derivative of g(x)=1+6x+3x^2 is 6x+6