What is (d^5)/(dx^5)(1/x+cos(2x)) ?

The answer to (d^5)/(dx^5)(1/x+cos(2x)) is -(120)/(x^6)-32sin(2x)

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(d^5)/(dx^5)(1/x+cos(2x))

\frac{d ^{5}}{d x ^{5}} (\frac{1}{x} + cos (2 x))

- \frac{120}{x ^{6}} - 32 sin (2 x)

Solution steps

\frac{d ^{5}}{d x ^{5}} (\frac{1}{x} + cos (2 x))

\frac{d}{d x} (\frac{1}{x} + cos (2 x)) = - \frac{1}{x ^{2}} - sin (2 x) \cdot 2

= \frac{d ^{4}}{d x ^{4}} (- \frac{1}{x ^{2}} - sin (2 x) \cdot 2)

\frac{d}{d x} (- \frac{1}{x ^{2}} - sin (2 x) \cdot 2) = \frac{2}{x ^{3}} - 4 cos (2 x)

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

\frac{d}{d x} (\frac{2}{x ^{3}} - 4 cos (2 x)) = - \frac{6}{x ^{4}} + 8 sin (2 x)

= \frac{d ^{2}}{d x ^{2}} (- \frac{6}{x ^{4}} + 8 sin (2 x))

\frac{d}{d x} (- \frac{6}{x ^{4}} + 8 sin (2 x)) = \frac{24}{x ^{5}} + 16 cos (2 x)

= \frac{d}{d x} (\frac{24}{x ^{5}} + 16 cos (2 x))

\frac{d}{d x} (\frac{24}{x ^{5}} + 16 cos (2 x)) = - \frac{120}{x ^{6}} - 32 sin (2 x)

= - \frac{120}{x ^{6}} - 32 sin (2 x)

What is (d^5)/(dx^5)(1/x+cos(2x)) ?
The answer to (d^5)/(dx^5)(1/x+cos(2x)) is -(120)/(x^6)-32sin(2x)