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

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

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

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

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

- 3 cos (3 x)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= - 3 cos (3 x) + 0

Simplify

= - 3 cos (3 x)

\frac{d}{d x} (\frac{4 x ^{4} + 7}{x ^{2}})

y^{^{'}} = 0.06 y - 0.0006 y^{2}

n = 1 \sum \infty \frac{n ^{3}}{n ^{3}}

\frac{d}{d x} (3 x^{3} - 2 x^{2} - 5 x + 2)

x \to 0 lim (x csc (3 x))

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