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}}

f (x) = ln (19 x^{3} + sin (3 x^{2}))

\frac{d}{d X} (X^{3} - 3 X)

x y^{^{'}} + y = - 8 x y^{2}, y (1) = 3

\frac{s + 1}{s ^{2} + 4 s + 8}

sin ((π x)^{8})