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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of x^2+8

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

2 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (8) = 0

= 2 x + 0

Simplify

= 2 x

(x + y) d x + x ln (x) d y = 0

\frac{\partial}{\partial y} (x e^{x y})

x \to 7 lim (tan (\frac{π x}{6}))

f (x) = 3 + 4 x^{2} - 2 x^{3}, (1, 5)

\frac{d ^{3} x}{d t ^{3}} = 4 t^{3} + t