What is the derivative of 0.01(0.2x^2+4x+64^{2/3}) ?

The derivative of 0.01(0.2x^2+4x+64^{2/3}) is (0.00666…(0.4x+4))/((0.2x^2+4x+64)^{1/3)}

What is the first derivative of 0.01(0.2x^2+4x+64^{2/3}) ?

The first derivative of 0.01(0.2x^2+4x+64^{2/3}) is (0.00666…(0.4x+4))/((0.2x^2+4x+64)^{1/3)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (0.01 (0.2 x^{2} + 4 x + 64)^{\frac{2}{3}})

\frac{0.00666 \dots ( 0.4 x + 4 )}{( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}}

Solution steps

\frac{d}{d x} (0.01 (0.2 x^{2} + 4 x + 64)^{\frac{2}{3}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.01 \frac{d}{d x} ((0.2 x^{2} + 4 x + 64)^{\frac{2}{3}})

Apply the chain rule:

\frac{2}{3 ( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}} \frac{d}{d x} (0.2 x^{2} + 4 x + 64)

= \frac{2}{3 ( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}} \frac{d}{d x} (0.2 x^{2} + 4 x + 64)

\frac{d}{d x} (0.2 x^{2} + 4 x + 64) = 0.4 x + 4

= 0.01 \cdot \frac{2}{3 ( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}} (0.4 x + 4)

Simplify

0.01 \cdot \frac{2}{3 ( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}} (0.4 x + 4) : \frac{0.00666 \dots ( 0.4 x + 4 )}{( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}}

= \frac{0.00666 \dots ( 0.4 x + 4 )}{( 0.2 x ^{2} + 4 x + 64 ) ^{\frac{1}{3}}}

\int_{0}^{10} x^{2} e^{- 0.08 x} d x

\frac{\partial}{\partial y} (y^{5} cos (3 x))

\int - \frac{x}{2} d x

\frac{( x ^{\frac{1}{3}} - 4 )}{x ^{\frac{7}{2}}}

\frac{d y}{d t} = t^{2}

What is the derivative of 0.01(0.2x^2+4x+64^{2/3}) ?
The derivative of 0.01(0.2x^2+4x+64^{2/3}) is (0.00666…(0.4x+4))/((0.2x^2+4x+64)^{1/3)}
What is the first derivative of 0.01(0.2x^2+4x+64^{2/3}) ?
The first derivative of 0.01(0.2x^2+4x+64^{2/3}) is (0.00666…(0.4x+4))/((0.2x^2+4x+64)^{1/3)}