What is the derivative of 4x^{x^3+3x} ?

The derivative of 4x^{x^3+3x} is 4x^{x^3+3x}(ln(x)(3x^2+3)+x^2+3)

What is the first derivative of 4x^{x^3+3x} ?

The first derivative of 4x^{x^3+3x} is 4x^{x^3+3x}(ln(x)(3x^2+3)+x^2+3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

4 x^{x^{3} + 3 x} (ln (x) (3 x^{2} + 3) + x^{2} + 3)

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (x^{x^{3} + 3 x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{x^{3} + 3 x} = e^{(x^{3} + 3 x) l n (x)}

= 4 \frac{d}{d x} (e^{(x^{3} + 3 x) l n (x)})

Apply the chain rule:

e^{(x^{3} + 3 x) l n (x)} \frac{d}{d x} ((x^{3} + 3 x) ln (x))

= e^{(x^{3} + 3 x) l n (x)} \frac{d}{d x} ((x^{3} + 3 x) ln (x))

\frac{d}{d x} ((x^{3} + 3 x) ln (x)) = ln (x) (3 x^{2} + 3) + x^{2} + 3

= 4 e^{(x^{3} + 3 x) l n (x)} (ln (x) (3 x^{2} + 3) + x^{2} + 3)

e^{(x^{3} + 3 x) l n (x)} = x^{x^{3} + 3 x}

= 4 x^{x^{3} + 3 x} (ln (x) (3 x^{2} + 3) + x^{2} + 3)

\int sin (x) cos^{5} (x) d x

V (r) = \frac{4}{3} π r^{3}

0.04 x^{2} - 2 x + 6 x^{0.2}

\frac{d}{d x} (sin (ln (x^{3})))

\int \frac{1}{x ^{3} - x ^{2} - 9 x + 9} d x

What is the derivative of 4x^{x^3+3x} ?
The derivative of 4x^{x^3+3x} is 4x^{x^3+3x}(ln(x)(3x^2+3)+x^2+3)
What is the first derivative of 4x^{x^3+3x} ?
The first derivative of 4x^{x^3+3x} is 4x^{x^3+3x}(ln(x)(3x^2+3)+x^2+3)