What is the d/(dy)(2y-1/x+cos(3x)) ?

The d/(dy)(2y-1/x+cos(3x)) is 2

What is the first d/(dy)(2y-1/x+cos(3x)) ?

The first d/(dy)(2y-1/x+cos(3x)) is 2

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\frac{d}{d y} (2 y - \frac{1}{x} + cos (3 x))

2

Solution steps

\frac{d}{d y} (2 y - \frac{1}{x} + cos (3 x))

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (2 y) - \frac{d}{d y} (\frac{1}{x}) + \frac{d}{d y} (cos (3 x))

\frac{d}{d y} (2 y) = 2

\frac{d}{d y} (\frac{1}{x}) = 0

\frac{d}{d y} (cos (3 x)) = 0

= 2 - 0 + 0

Simplify

= 2

\int (x \cdot cos (5 x)) d x

e^{\frac{1}{x}} 0

x \to + 0 lim (\frac{2 ^{x} - 3 ^{x}}{x})

e^{9 e^{x}}

What is the d/(dy)(2y-1/x+cos(3x)) ?
The d/(dy)(2y-1/x+cos(3x)) is 2
What is the first d/(dy)(2y-1/x+cos(3x)) ?
The first d/(dy)(2y-1/x+cos(3x)) is 2