What is the derivative of (((x^3+8)/(2x))^7) ?

The derivative of (((x^3+8)/(2x))^7) is (7(x^3+8)^6(x^3-4))/(64x^8)

What is the first derivative of (((x^3+8)/(2x))^7) ?

The first derivative of (((x^3+8)/(2x))^7) is (7(x^3+8)^6(x^3-4))/(64x^8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{( x ^{3} + 8 )}{2 x})^{7}

\frac{7 ( x ^{3} + 8 ) ^{6} ( x ^{3} - 4 )}{64 x ^{8}}

Solution steps

\frac{d}{d x} (\frac{( x ^{3} + 8 )}{2 x})^{7}

Apply the chain rule:

7 (\frac{x ^{3} + 8}{2 x})^{6} \frac{d}{d x} (\frac{x ^{3} + 8}{2 x})

= 7 (\frac{x ^{3} + 8}{2 x})^{6} \frac{d}{d x} (\frac{x ^{3} + 8}{2 x})

\frac{d}{d x} (\frac{x ^{3} + 8}{2 x}) = \frac{x ^{3} - 4}{x ^{2}}

= 7 (\frac{x ^{3} + 8}{2 x})^{6} \frac{x ^{3} - 4}{x ^{2}}

Simplify

7 (\frac{x ^{3} + 8}{2 x})^{6} \frac{x ^{3} - 4}{x ^{2}} : \frac{7 ( x ^{3} + 8 ) ^{6} ( x ^{3} - 4 )}{64 x ^{8}}

= \frac{7 ( x ^{3} + 8 ) ^{6} ( x ^{3} - 4 )}{64 x ^{8}}

u^{6}

x \to 0 lim ((\frac{1}{x})^{x})

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

n = 2 \sum \infty \frac{n + 4}{n !}

\frac{x ^{2} - 1}{x ^{2} + 1}

What is the derivative of (((x^3+8)/(2x))^7) ?
The derivative of (((x^3+8)/(2x))^7) is (7(x^3+8)^6(x^3-4))/(64x^8)
What is the first derivative of (((x^3+8)/(2x))^7) ?
The first derivative of (((x^3+8)/(2x))^7) is (7(x^3+8)^6(x^3-4))/(64x^8)