What is the derivative of 1+3x+9x^2 ?

The derivative of 1+3x+9x^2 is 18x+3

What is the first derivative of 1+3x+9x^2 ?

The first derivative of 1+3x+9x^2 is 18x+3

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\frac{d}{d x} (1 + 3 x + 9 x^{2})

18 x + 3

Solution steps

\frac{d}{d x} (1 + 3 x + 9 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1) + \frac{d}{d x} (3 x) + \frac{d}{d x} (9 x^{2})

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

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

\frac{d}{d x} (9 x^{2}) = 18 x

= 0 + 3 + 18 x

Simplify

= 18 x + 3

f (x) = 5 x^{2}

(9 x^{2} + y - 1) d x - (14 y - x) d y = 0, y (1) = 0

\int_{0}^{1} (x + 1 - x^{2}) d x

\int 1 0^{x} (2^{- x} + 5^{- x}) d x

\frac{d}{d x} (A x^{2} e^{2 x})

What is the derivative of 1+3x+9x^2 ?
The derivative of 1+3x+9x^2 is 18x+3
What is the first derivative of 1+3x+9x^2 ?
The first derivative of 1+3x+9x^2 is 18x+3