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

The answer to (\partial)/(\partial x)(3x^2+ycos(xy)) is 6x-y^2sin(yx)

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(\partial)/(\partial x)(3x^2+ycos(xy))

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

6 x - y^{2} sin (y x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

\frac{\partial}{\partial x} (y cos (x y)) = - y^{2} sin (y x)

= 6 x - y^{2} sin (y x)

\frac{\partial}{\partial x} (sin (2 x) cos (8 y))

\frac{d ^{2}}{d x ^{2}} (x^{2} + 21)

\frac{d}{d u} (\frac{1}{u})

\int_{0}^{3} (2 x - x^{2})^{4} (1 - x) d x

\frac{d}{d x} (\frac{1}{4} (x^{2} - 4))

What is (\partial)/(\partial x)(3x^2+ycos(xy)) ?
The answer to (\partial)/(\partial x)(3x^2+ycos(xy)) is 6x-y^2sin(yx)