What is (d^4)/(dx^4)((8x)/(1-x)) ?

The answer to (d^4)/(dx^4)((8x)/(1-x)) is (192)/((1-x)^5)

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(d^4)/(dx^4)((8x)/(1-x))

\frac{d ^{4}}{d x ^{4}} (\frac{8 x}{1 - x})

\frac{192}{( 1 - x ) ^{5}}

Solution steps

\frac{d ^{4}}{d x ^{4}} (\frac{8 x}{1 - x})

\frac{d}{d x} (\frac{8 x}{1 - x}) = \frac{8}{( 1 - x ) ^{2}}

= \frac{d ^{3}}{d x ^{3}} (\frac{8}{( 1 - x ) ^{2}})

\frac{d}{d x} (\frac{8}{( 1 - x ) ^{2}}) = \frac{16}{( 1 - x ) ^{3}}

= \frac{d ^{2}}{d x ^{2}} (\frac{16}{( 1 - x ) ^{3}})

\frac{d}{d x} (\frac{16}{( 1 - x ) ^{3}}) = \frac{48}{( 1 - x ) ^{4}}

= \frac{d}{d x} (\frac{48}{( 1 - x ) ^{4}})

\frac{d}{d x} (\frac{48}{( 1 - x ) ^{4}}) = \frac{192}{( 1 - x ) ^{5}}

= \frac{192}{( 1 - x ) ^{5}}

f (x) = 5^{5 x + 1} + 3^{1 - 3 x}

y^{^{'}} + 5 y = 7

\frac{d ^{2}}{d x ^{2}} (- 2 x^{4} - 2 x^{3} + 60 x^{2} - 22)

\frac{d}{d x} (2 x cos (x) - 2 sin (x))

\frac{x}{1 + x ^{2}}, at x = 2

What is (d^4)/(dx^4)((8x)/(1-x)) ?
The answer to (d^4)/(dx^4)((8x)/(1-x)) is (192)/((1-x)^5)