What is the derivative of y=((8x^2-5)/(2-x))^3 ?

The derivative of y=((8x^2-5)/(2-x))^3 is (3(8x^2-5)^2(32x-8x^2-5))/((2-x)^4)

What is the first derivative of y=((8x^2-5)/(2-x))^3 ?

The first derivative of y=((8x^2-5)/(2-x))^3 is (3(8x^2-5)^2(32x-8x^2-5))/((2-x)^4)

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derivative of

y = (\frac{8 x ^{2} - 5}{2 - x})^{3}

\frac{3 ( 8 x ^{2} - 5 ) ^{2} ( 32 x - 8 x ^{2} - 5 )}{( 2 - x ) ^{4}}

Solution steps

\frac{d}{d x} ((\frac{8 x ^{2} - 5}{2 - x})^{3})

Apply the chain rule:

3 (\frac{8 x ^{2} - 5}{2 - x})^{2} \frac{d}{d x} (\frac{8 x ^{2} - 5}{2 - x})

= 3 (\frac{8 x ^{2} - 5}{2 - x})^{2} \frac{d}{d x} (\frac{8 x ^{2} - 5}{2 - x})

\frac{d}{d x} (\frac{8 x ^{2} - 5}{2 - x}) = \frac{32 x - 8 x ^{2} - 5}{( 2 - x ) ^{2}}

= 3 (\frac{8 x ^{2} - 5}{2 - x})^{2} \frac{32 x - 8 x ^{2} - 5}{( 2 - x ) ^{2}}

Simplify

3 (\frac{8 x ^{2} - 5}{2 - x})^{2} \frac{32 x - 8 x ^{2} - 5}{( 2 - x ) ^{2}} : \frac{3 ( 8 x ^{2} - 5 ) ^{2} ( 32 x - 8 x ^{2} - 5 )}{( 2 - x ) ^{4}}

= \frac{3 ( 8 x ^{2} - 5 ) ^{2} ( 32 x - 8 x ^{2} - 5 )}{( 2 - x ) ^{4}}

g (x) = lo g_{7} (8 x^{3} - 9 x^{2})

\int x (x sin (x) - cos (x)) d x

\int_{0}^{0.9} \frac{1}{( 1 - x )} d x

\int 6 cos (5 x) d x

y = 2 (x + y) y^{^{'}}

What is the derivative of y=((8x^2-5)/(2-x))^3 ?
The derivative of y=((8x^2-5)/(2-x))^3 is (3(8x^2-5)^2(32x-8x^2-5))/((2-x)^4)
What is the first derivative of y=((8x^2-5)/(2-x))^3 ?
The first derivative of y=((8x^2-5)/(2-x))^3 is (3(8x^2-5)^2(32x-8x^2-5))/((2-x)^4)