What is (\partial)/(\partial x)(3y^2+xcos(xy)) ?

The answer to (\partial)/(\partial x)(3y^2+xcos(xy)) is cos(yx)-yxsin(yx)

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(\partial)/(\partial x)(3y^2+xcos(xy))

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

cos (y x) - y x sin (y x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

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

= 0 + cos (y x) - y x sin (y x)

Simplify

= cos (y x) - y x sin (y x)

\int \frac{x ^{3} + 1}{x + 1} d x

x \to 3 - lim (\frac{3}{x - 3})

\frac{d}{d x} (2 x - 2 - 2 cos (x - 2))

3 x + 2

\frac{d ^{2}}{d x ^{2}} (\frac{3}{4 t ^{6}})

What is (\partial)/(\partial x)(3y^2+xcos(xy)) ?
The answer to (\partial)/(\partial x)(3y^2+xcos(xy)) is cos(yx)-yxsin(yx)