What is the derivative of \sqrt[5]{(x^3-2/(x^3+2)}) ?

The derivative of \sqrt[5]{(x^3-2/(x^3+2)}) is (12x^2)/(5(x^3-2)^{4/5)(x^3+2)^{6/5}}

What is the first derivative of \sqrt[5]{(x^3-2/(x^3+2)}) ?

The first derivative of \sqrt[5]{(x^3-2/(x^3+2)}) is (12x^2)/(5(x^3-2)^{4/5)(x^3+2)^{6/5}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{12 x ^{2}}{5 ( x ^{3} - 2 ) ^{\frac{4}{5}} ( x ^{3} + 2 ) ^{\frac{6}{5}}}

Solution steps

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{x ^{3} - 2}{x ^{3} + 2}) = \frac{12 x ^{2}}{( x ^{3} + 2 ) ^{2}}

= \frac{1}{5 ( \frac{x ^{3} - 2}{x ^{3} + 2} ) ^{\frac{4}{5}}} \cdot \frac{12 x ^{2}}{( x ^{3} + 2 ) ^{2}}

Simplify

\frac{1}{5 ( \frac{x ^{3} - 2}{x ^{3} + 2} ) ^{\frac{4}{5}}} \cdot \frac{12 x ^{2}}{( x ^{3} + 2 ) ^{2}} : \frac{12 x ^{2}}{5 ( x ^{3} - 2 ) ^{\frac{4}{5}} ( x ^{3} + 2 ) ^{\frac{6}{5}}}

= \frac{12 x ^{2}}{5 ( x ^{3} - 2 ) ^{\frac{4}{5}} ( x ^{3} + 2 ) ^{\frac{6}{5}}}

y^{^{'}} = x e^{6}

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

(1 + 9 x^{2}) (x + x^{2})

\int \frac{2}{4 - 9 x ^{2}} d x

x \to - 4 lim (6 - 3 x)

What is the derivative of \sqrt[5]{(x^3-2/(x^3+2)}) ?
The derivative of \sqrt[5]{(x^3-2/(x^3+2)}) is (12x^2)/(5(x^3-2)^{4/5)(x^3+2)^{6/5}}
What is the first derivative of \sqrt[5]{(x^3-2/(x^3+2)}) ?
The first derivative of \sqrt[5]{(x^3-2/(x^3+2)}) is (12x^2)/(5(x^3-2)^{4/5)(x^3+2)^{6/5}}