What is the derivative of x/(\sqrt[3]{x^2+4}) ?

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

Solution steps

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

\frac{d x}{d x} = 1

Simplify

f (t) = t

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

\int \frac{12}{x ^{2} + 1} d x

(1 - x)^{\frac{1}{5}}

\frac{d}{d x} (\frac{x}{1 + x})

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