What is the first derivative of x^2+64 ?

The first derivative of x^2+64 is 2x

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derivative of x^2+64

derivative of

x^{2} + 64

2 x

Solution steps

\frac{d}{d x} (x^{2} + 64)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (64)

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

\frac{d}{d x} (64) = 0

= 2 x + 0

Simplify

= 2 x

\frac{\partial}{\partial x} (9 e^{x y + 8})

x = y^{2} - 10, x = 3 y

c y^{- 2}

n = 0 \sum \infty \frac{1}{2 + 6 ^{- n}}

\frac{d}{d x} (\frac{1}{25 - x ^{2}})