What is (\partial)/(\partial x)(xycos(3x)+e^{xy}) ?

The answer to (\partial)/(\partial x)(xycos(3x)+e^{xy}) is y(cos(3x)-3xsin(3x))+e^{xy}y

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(\partial)/(\partial x)(xycos(3x)+e^{xy})

\frac{\partial}{\partial x} (x y cos (3 x) + e^{x y})

y (cos (3 x) - 3 x sin (3 x)) + e^{x y} y

Solution steps

\frac{\partial}{\partial x} (x y cos (3 x) + e^{x y})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x y cos (3 x)) + \frac{\partial}{\partial x} (e^{x y})

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

\frac{\partial}{\partial x} (e^{x y}) = e^{x y} y

= y (cos (3 x) - 3 x sin (3 x)) + e^{x y} y

(x + 1) y^{^{'}} - y = x

\int (2 x^{5}) d x

\frac{x d y}{d x} + y = - 8 x y^{2}, y (1) = 5

f (x) = \frac{x}{4} arcsin (x)

x y^{^{'}} + y = x^{3}

What is (\partial)/(\partial x)(xycos(3x)+e^{xy}) ?
The answer to (\partial)/(\partial x)(xycos(3x)+e^{xy}) is y(cos(3x)-3xsin(3x))+e^{xy}y