What is the derivative of y= 1/x+cos(2x) ?

The derivative of y= 1/x+cos(2x) is -1/(x^2)-sin(2x)*2

What is the first derivative of y= 1/x+cos(2x) ?

The first derivative of y= 1/x+cos(2x) is -1/(x^2)-sin(2x)*2

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derivative of

y = \frac{1}{x} + cos (2 x)

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

x \to 4 + lim (16 - x^{2})

\int_{0}^{1} 6 e^{2 x} d x

\frac{d}{d x} ((x^{4} + 3 x)^{- 1})

\int \frac{1}{x ( 1 + 4 x ^{2} )} d x

What is the derivative of y= 1/x+cos(2x) ?
The derivative of y= 1/x+cos(2x) is -1/(x^2)-sin(2x)*2
What is the first derivative of y= 1/x+cos(2x) ?
The first derivative of y= 1/x+cos(2x) is -1/(x^2)-sin(2x)*2