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

The derivative of y=((8x^2)/(2-x))^3 is (1536x^4(4x-x^2))/((2-x)^4)

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

The first derivative of y=((8x^2)/(2-x))^3 is (1536x^4(4x-x^2))/((2-x)^4)

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derivative of

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

\frac{1536 x ^{4} ( 4 x - x ^{2} )}{( 2 - x ) ^{4}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

3 (\frac{8 x ^{2}}{2 - x})^{2} \frac{8 ( 4 x - x ^{2} )}{( 2 - x ) ^{2}} : \frac{1536 x ^{4} ( 4 x - x ^{2} )}{( 2 - x ) ^{4}}

= \frac{1536 x ^{4} ( 4 x - x ^{2} )}{( 2 - x ) ^{4}}

cos (x), sin (x), [0, \frac{π}{2}]

\int x^{5} 9 x^{2} - 1 d x

\frac{\partial}{\partial y} (3 x^{2} y + cos (y))

t y^{^{'}} + y t = 1 - y

\int_{0}^{π} sec^{2} (\frac{x}{4}) d x

What is the derivative of y=((8x^2)/(2-x))^3 ?
The derivative of y=((8x^2)/(2-x))^3 is (1536x^4(4x-x^2))/((2-x)^4)
What is the first derivative of y=((8x^2)/(2-x))^3 ?
The first derivative of y=((8x^2)/(2-x))^3 is (1536x^4(4x-x^2))/((2-x)^4)