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

The derivative of f(x)=((4x^2)/(2-x))^3 is (192x^4(4x-x^2))/((2-x)^4)

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

The first derivative of f(x)=((4x^2)/(2-x))^3 is (192x^4(4x-x^2))/((2-x)^4)

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derivative of

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

\frac{192 x ^{4} ( 4 x - x ^{2} )}{( 2 - x ) ^{4}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

3 (\frac{4 x ^{2}}{2 - x})^{2} \frac{4 ( 4 x - x ^{2} )}{( 2 - x ) ^{2}} : \frac{192 x ^{4} ( 4 x - x ^{2} )}{( 2 - x ) ^{4}}

= \frac{192 x ^{4} ( 4 x - x ^{2} )}{( 2 - x ) ^{4}}

\frac{11 - 3}{p - 1}

\frac{\partial}{\partial x} (y^{\frac{x}{z}})

f (x) = (2 x - x) (5 x + 2)

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

x y^{^{'}} + y + y^{2} = 0

What is the derivative of f(x)=((4x^2)/(2-x))^3 ?
The derivative of f(x)=((4x^2)/(2-x))^3 is (192x^4(4x-x^2))/((2-x)^4)
What is the first derivative of f(x)=((4x^2)/(2-x))^3 ?
The first derivative of f(x)=((4x^2)/(2-x))^3 is (192x^4(4x-x^2))/((2-x)^4)