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

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

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

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

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

Solution steps

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

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

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

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

\int 4 sec^{4} (x) tan (x) d x

\frac{d}{d x} (\frac{x ^{5}}{5} + \frac{1}{12 x ^{3}})

x d y - (x^{2} - y^{2} + y) \cdot d x = 0

y = \frac{x + 2}{x - 2}, (3, 5)

\int \frac{8 x ^{3} + 3}{( x ^{2} + 2 ) ^{2}} d x

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