What is the derivative of x^3ln(x^4+1) ?

The derivative of x^3ln(x^4+1) is 3x^2ln(x^4+1)+(4x^6)/(x^4+1)

What is the first derivative of x^3ln(x^4+1) ?

The first derivative of x^3ln(x^4+1) is 3x^2ln(x^4+1)+(4x^6)/(x^4+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (x^{3} ln (x^{4} + 1))

3 x^{2} ln (x^{4} + 1) + \frac{4 x ^{6}}{x ^{4} + 1}

Solution steps

\frac{d}{d x} (x^{3} ln (x^{4} + 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3}, g = ln (x^{4} + 1)

= \frac{d}{d x} (x^{3}) ln (x^{4} + 1) + \frac{d}{d x} (ln (x^{4} + 1)) x^{3}

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

\frac{d}{d x} (ln (x^{4} + 1)) = \frac{4 x ^{3}}{x ^{4} + 1}

= 3 x^{2} ln (x^{4} + 1) + \frac{4 x ^{3}}{x ^{4} + 1} x^{3}

\frac{4 x ^{3}}{x ^{4} + 1} x^{3} = \frac{4 x ^{6}}{x ^{4} + 1}

= 3 x^{2} ln (x^{4} + 1) + \frac{4 x ^{6}}{x ^{4} + 1}

\int_{2}^{x^{2}} cos (t^{2} + 1) d t

d er i v a t i v e y = x cos (x)

a re a x = 63 - 7 y^{2}, x = 7 y^{2} - 63

d er i v a t i v e t^{3} - t

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

What is the derivative of x^3ln(x^4+1) ?
The derivative of x^3ln(x^4+1) is 3x^2ln(x^4+1)+(4x^6)/(x^4+1)
What is the first derivative of x^3ln(x^4+1) ?
The first derivative of x^3ln(x^4+1) is 3x^2ln(x^4+1)+(4x^6)/(x^4+1)