What is the derivative of ((6x^2/(4-x))^3) ?

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

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

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

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\frac{d}{d x} ((\frac{6 x ^{2}}{4 - x})^{3})

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

Solution steps

\frac{d}{d x} ((\frac{6 x ^{2}}{4 - x})^{3})

Apply the chain rule:

3 (\frac{6 x ^{2}}{4 - x})^{2} \frac{d}{d x} (\frac{6 x ^{2}}{4 - x})

= 3 (\frac{6 x ^{2}}{4 - x})^{2} \frac{d}{d x} (\frac{6 x ^{2}}{4 - x})

\frac{d}{d x} (\frac{6 x ^{2}}{4 - x}) = \frac{6 ( 8 x - x ^{2} )}{( 4 - x ) ^{2}}

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

Simplify

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

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

n = 0 \sum \infty \frac{n}{n ^{4} + 1}

\frac{d}{d x} (\frac{x + 9}{x ^{3} + x - 8})

\int_{- 1}^{3} \frac{x}{x ^{2} + 1} d x

- 6 cos (t)

y = 2 x^{3}, (- 2, - 16)

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