What is the derivative of ((6x^2)/(8-x))^3 ?

The derivative of ((6x^2)/(8-x))^3 is (648x^4(16x-x^2))/((8-x)^4)

What is the first derivative of ((6x^2)/(8-x))^3 ?

The first derivative of ((6x^2)/(8-x))^3 is (648x^4(16x-x^2))/((8-x)^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

(\frac{6 x ^{2}}{8 - x})^{3}

\frac{648 x ^{4} ( 16 x - x ^{2} )}{( 8 - x ) ^{4}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{6 x ^{2}}{8 - x}) = \frac{6 ( 16 x - x ^{2} )}{( 8 - x ) ^{2}}

= 3 (\frac{6 x ^{2}}{8 - x})^{2} \frac{6 ( 16 x - x ^{2} )}{( 8 - x ) ^{2}}

Simplify

3 (\frac{6 x ^{2}}{8 - x})^{2} \frac{6 ( 16 x - x ^{2} )}{( 8 - x ) ^{2}} : \frac{648 x ^{4} ( 16 x - x ^{2} )}{( 8 - x ) ^{4}}

= \frac{648 x ^{4} ( 16 x - x ^{2} )}{( 8 - x ) ^{4}}

x^{- 5}

\int 4 sin (5 x) d x

\int \frac{1}{x ^{\frac{2}{3}} ( 8 + x ^{\frac{1}{3}} )} d x

y (θ) = ln (θ + 1) - e^{θ}

\frac{\partial}{\partial x} (ln (x + y) - 5 y)

What is the derivative of ((6x^2)/(8-x))^3 ?
The derivative of ((6x^2)/(8-x))^3 is (648x^4(16x-x^2))/((8-x)^4)
What is the first derivative of ((6x^2)/(8-x))^3 ?
The first derivative of ((6x^2)/(8-x))^3 is (648x^4(16x-x^2))/((8-x)^4)