What is (\partial)/(\partial y)(y^2-cos(xy)) ?

The answer to (\partial)/(\partial y)(y^2-cos(xy)) is 2y+xsin(yx)

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(\partial)/(\partial y)(y^2-cos(xy))

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

2 y + x sin (y x)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 y - (- sin (x y) x)

Simplify

= 2 y + x sin (y x)

x \to 0.99 lim (2 x^{2} - 5 x + 1)

\int 2 tan (x) d x

\int (1 - cot (x)) csc^{2} (x) d x

\frac{d}{d θ} (- 6 sin (θ))

\frac{d}{d x} (- \frac{1}{( 2 x + 8 ) ^{\frac{3}{2}}})

What is (\partial)/(\partial y)(y^2-cos(xy)) ?
The answer to (\partial)/(\partial y)(y^2-cos(xy)) is 2y+xsin(yx)