What is the (e^{x^2+1})^' ?

The (e^{x^2+1})^' is e^{x^2+1}*2x

What is the first (e^{x^2+1})^' ?

The first (e^{x^2+1})^' is e^{x^2+1}*2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(e^{x^{2} + 1})^{^{'}}

e^{x^{2} + 1} \cdot 2 x

Solution steps

(e^{x^{2} + 1})^{'}

Apply the chain rule:

e^{x^{2} + 1} (x^{2} + 1)^{'}

= e^{x^{2} + 1} (x^{2} + 1)^{'}

(x^{2} + 1)^{'} = 2 x

= e^{x^{2} + 1} \cdot 2 x

x^{3} os (x)

11 sin (t) + 14 cos (t)

\frac{d}{d x} (x^{2} + 4 x + 12 y - 8)

\frac{\partial}{\partial x} (x^{2} y^{2} + 4 x^{2} - 3 x y)

f^{^{''}} (x) = a cos (2 x) + b sin (2 x)