What is (\partial)/(\partial x)(x/y+sin(2x+3y)) ?

The answer to (\partial)/(\partial x)(x/y+sin(2x+3y)) is 1/y+cos(2x+3y)*2

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(\partial)/(\partial x)(x/y+sin(2x+3y))

\frac{\partial}{\partial x} (\frac{x}{y} + sin (2 x + 3 y))

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

Solution steps

\frac{\partial}{\partial x} (\frac{x}{y} + sin (2 x + 3 y))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (\frac{x}{y}) + \frac{\partial}{\partial x} (sin (2 x + 3 y))

\frac{\partial}{\partial x} (\frac{x}{y}) = \frac{1}{y}

\frac{\partial}{\partial x} (sin (2 x + 3 y)) = cos (2 x + 3 y) \cdot 2

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

\int_{0}^{6} \frac{1}{u ^{2} - 36} d u

x \to 2 lim (\frac{∣ x - 2 ∣}{x - 2})

\int b cos (x) d x

\frac{d y}{d x} = \frac{y}{x} + 9 x + 4

\frac{\partial}{\partial x} (9 - x)

What is (\partial)/(\partial x)(x/y+sin(2x+3y)) ?
The answer to (\partial)/(\partial x)(x/y+sin(2x+3y)) is 1/y+cos(2x+3y)*2