What is the derivative of y=(x^2+1) ?

The derivative of y=(x^2+1) is 2x

What is the first derivative of y=(x^2+1) ?

The first derivative of y=(x^2+1) is 2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = (x^{2} + 1)

2 x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 2 x + 0

Simplify

= 2 x

\int \frac{36 + ( ln ( 8 x ) ) ^{2}}{x} d x

x \to 0 lim (\frac{x ^{2} - 9}{x + 3})

y^{^{'}} = \frac{( x + y + 1 )}{( 2 x + 2 y + 3 )}

\frac{d}{d x} (\frac{4 x - 3}{5 x ^{2} + 3 x})

\int sec^{5} (\frac{θ}{2}) d θ

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