What is the derivative of-(cos(1/x))/(x^2) ?

The derivative of-(cos(1/x))/(x^2) is -(sin(1/x)-2xcos(1/x))/(x^4)

What is the first derivative of-(cos(1/x))/(x^2) ?

The first derivative of-(cos(1/x))/(x^2) is -(sin(1/x)-2xcos(1/x))/(x^4)

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derivative of

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

- \frac{sin ( \frac{1}{x} ) - 2 x cos ( \frac{1}{x} )}{x ^{4}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d}{d x} (x^{2}) = 2 x

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

Simplify

- \frac{\frac{s i n ( \frac{1}{x} )}{x ^{2}} x ^{2} - 2 x cos ( \frac{1}{x} )}{( x ^{2} ) ^{2}} : - \frac{sin ( \frac{1}{x} ) - 2 x cos ( \frac{1}{x} )}{x ^{4}}

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

y = e^{x}, (0, 1)

\frac{6}{3 x + 1}

\frac{d}{d x} (\frac{3 x ^{5} + 9}{x ^{3}})

\int \frac{1}{8 x + 1} d x

x \to 1 + lim (ln (x - 1))

What is the derivative of-(cos(1/x))/(x^2) ?
The derivative of-(cos(1/x))/(x^2) is -(sin(1/x)-2xcos(1/x))/(x^4)
What is the first derivative of-(cos(1/x))/(x^2) ?
The first derivative of-(cos(1/x))/(x^2) is -(sin(1/x)-2xcos(1/x))/(x^4)