What is the derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) ?

The derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) is (16x^3+15x^2)/(3(x^3)^{2/3)(4x+5)^{2/3}}

What is the first derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) ?

The first derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) is (16x^3+15x^2)/(3(x^3)^{2/3)(4x+5)^{2/3}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{16 x ^{3} + 15 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 4 x + 5 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (4 x^{4} + 5 x^{3})

= \frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (4 x^{4} + 5 x^{3})

\frac{d}{d x} (4 x^{4} + 5 x^{3}) = 16 x^{3} + 15 x^{2}

= \frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} (16 x^{3} + 15 x^{2})

Simplify

\frac{1}{3 ( 4 x ^{4} + 5 x ^{3} ) ^{\frac{2}{3}}} (16 x^{3} + 15 x^{2}) : \frac{16 x ^{3} + 15 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 4 x + 5 ) ^{\frac{2}{3}}}

= \frac{16 x ^{3} + 15 x ^{2}}{3 ( x ^{3} ) ^{\frac{2}{3}} ( 4 x + 5 ) ^{\frac{2}{3}}}

\int_{2}^{4} \frac{6 - x - y}{8} d y

n = 1 \sum \infty \frac{n}{2 ^{n} + 5}

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

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

\frac{2}{s + 1} + \frac{8 s}{s ^{2} + 9}

What is the derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) ?
The derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) is (16x^3+15x^2)/(3(x^3)^{2/3)(4x+5)^{2/3}}
What is the first derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) ?
The first derivative of f(x)=(\sqrt[3]{4x^4+5x^3}) is (16x^3+15x^2)/(3(x^3)^{2/3)(4x+5)^{2/3}}