What is the (cos(1/x))^' ?

The (cos(1/x))^' is (sin(1/x))/(x^2)

What is the first (cos(1/x))^' ?

The first (cos(1/x))^' is (sin(1/x))/(x^2)

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(cos (\frac{1}{x}))^{^{'}}

\frac{sin ( \frac{1}{x} )}{x ^{2}}

Solution steps

(cos (\frac{1}{x}))^{'}

Apply the chain rule:

- sin (\frac{1}{x}) (\frac{1}{x})^{'}

= - sin (\frac{1}{x}) (\frac{1}{x})^{'}

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

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

Simplify

- sin (\frac{1}{x}) (- \frac{1}{x ^{2}}) : \frac{sin ( \frac{1}{x} )}{x ^{2}}

= \frac{sin ( \frac{1}{x} )}{x ^{2}}

\int \frac{30 x ^{2} - 49}{x ^{4}} d x

\frac{\partial}{\partial x} (\frac{x}{x - 1})

x \to 0 lim (2 x^{2} cos (\frac{5}{x}))

f (x) = 5 x^{5} ln (x)

\frac{d y}{d x} = \frac{y}{x} + 5 x + 4