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

The answer to (\partial)/(\partial x)((x^2-2x)cos(y)) is cos(y)(2x-2)

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

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

cos (y) (2 x - 2)

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

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

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

= cos (y) (2 x - 2)

\frac{d}{d x} ((e^{3 x}) (cos (2 x)))

16 y^{''} + 8 y^{'} + y = 0

x \to 0 lim (\frac{tan ( 4 x ) cos ( 6 x ) - tan ( 4 x )}{24 x ^{3}})

x \to + 1 lim (\frac{x - 1}{x ^{6} - 1})

x \to 0 lim (\frac{3 x}{sin ( 10 x )})

What is (\partial)/(\partial x)((x^2-2x)cos(y)) ?
The answer to (\partial)/(\partial x)((x^2-2x)cos(y)) is cos(y)(2x-2)