What is the first derivative of 4x^2+9 ?

The first derivative of 4x^2+9 is 8x

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derivative of 4x^2+9

derivative of

4 x^{2} + 9

8 x

Solution steps

\frac{d}{d x} (4 x^{2} + 9)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (9)

\frac{d}{d x} (4 x^{2}) = 8 x

\frac{d}{d x} (9) = 0

= 8 x + 0

Simplify

= 8 x

\int (7 x + cos (x)) d x

\frac{d y}{d x} + \frac{2}{x} y = \frac{1}{x ^{2}}

n \to \infty lim (\frac{2}{n})

f (x) = \frac{3 ln ( x )}{13 x ^{2}}

x n \to 2 lim ((x n)^{2} - 5 x n + 3)