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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

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

Simplify

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

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

\int tan^{3} (2 x) d x

y = f (x), at x = 7

5^{7 - x^{2}}

8 t

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