What is the derivative of y=((4x^2-5)/(2-x))^3 ?

The derivative of y=((4x^2-5)/(2-x))^3 is (3(4x^2-5)^2(16x-4x^2-5))/((2-x)^4)

What is the first derivative of y=((4x^2-5)/(2-x))^3 ?

The first derivative of y=((4x^2-5)/(2-x))^3 is (3(4x^2-5)^2(16x-4x^2-5))/((2-x)^4)

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derivative of

y = (\frac{4 x ^{2} - 5}{2 - x})^{3}

\frac{3 ( 4 x ^{2} - 5 ) ^{2} ( 16 x - 4 x ^{2} - 5 )}{( 2 - x ) ^{4}}

Solution steps

\frac{d}{d x} ((\frac{4 x ^{2} - 5}{2 - x})^{3})

Apply the chain rule:

3 (\frac{4 x ^{2} - 5}{2 - x})^{2} \frac{d}{d x} (\frac{4 x ^{2} - 5}{2 - x})

= 3 (\frac{4 x ^{2} - 5}{2 - x})^{2} \frac{d}{d x} (\frac{4 x ^{2} - 5}{2 - x})

\frac{d}{d x} (\frac{4 x ^{2} - 5}{2 - x}) = \frac{16 x - 4 x ^{2} - 5}{( 2 - x ) ^{2}}

= 3 (\frac{4 x ^{2} - 5}{2 - x})^{2} \frac{16 x - 4 x ^{2} - 5}{( 2 - x ) ^{2}}

Simplify

3 (\frac{4 x ^{2} - 5}{2 - x})^{2} \frac{16 x - 4 x ^{2} - 5}{( 2 - x ) ^{2}} : \frac{3 ( 4 x ^{2} - 5 ) ^{2} ( 16 x - 4 x ^{2} - 5 )}{( 2 - x ) ^{4}}

= \frac{3 ( 4 x ^{2} - 5 ) ^{2} ( 16 x - 4 x ^{2} - 5 )}{( 2 - x ) ^{4}}

y^{^{'}} - \frac{2}{x} y = \frac{y ^{3}}{x ^{2}}, y (1) = 1

\frac{d}{d s} (\frac{1}{s ^{2} - 1})

\int \frac{csc ( θ )}{csc ( θ ) - sin ( θ )} d θ

\int 5 e^{- 2 t} sin (t) d t

\int (x + 2)^{2} (2 - x)^{6} d x

What is the derivative of y=((4x^2-5)/(2-x))^3 ?
The derivative of y=((4x^2-5)/(2-x))^3 is (3(4x^2-5)^2(16x-4x^2-5))/((2-x)^4)
What is the first derivative of y=((4x^2-5)/(2-x))^3 ?
The first derivative of y=((4x^2-5)/(2-x))^3 is (3(4x^2-5)^2(16x-4x^2-5))/((2-x)^4)