What is the derivative of \sqrt[3]{6x^2+7x+2} ?

The derivative of \sqrt[3]{6x^2+7x+2} is (12x+7)/(3(6x^2+7x+2)^{2/3)}

What is the first derivative of \sqrt[3]{6x^2+7x+2} ?

The first derivative of \sqrt[3]{6x^2+7x+2} is (12x+7)/(3(6x^2+7x+2)^{2/3)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{12 x + 7}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}} \frac{d}{d x} (6 x^{2} + 7 x + 2)

= \frac{1}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}} \frac{d}{d x} (6 x^{2} + 7 x + 2)

\frac{d}{d x} (6 x^{2} + 7 x + 2) = 12 x + 7

= \frac{1}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}} (12 x + 7)

Simplify

\frac{1}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}} (12 x + 7) : \frac{12 x + 7}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}}

= \frac{12 x + 7}{3 ( 6 x ^{2} + 7 x + 2 ) ^{\frac{2}{3}}}

x \to 0 lim (5 - \frac{∣ x ∣}{x})

\int_{- π}^{π} cos (x) d x

(- 9, - 6), (- 4, 5)

\frac{\partial}{\partial v} (u^{2} + v^{2} - 4)

x y^{^{'}} + y = - 7 x y^{2}, y (1) = - 6

What is the derivative of \sqrt[3]{6x^2+7x+2} ?
The derivative of \sqrt[3]{6x^2+7x+2} is (12x+7)/(3(6x^2+7x+2)^{2/3)}
What is the first derivative of \sqrt[3]{6x^2+7x+2} ?
The first derivative of \sqrt[3]{6x^2+7x+2} is (12x+7)/(3(6x^2+7x+2)^{2/3)}