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

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

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

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

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

- \frac{cos ( \frac{1}{x} )}{x ^{2}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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