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

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

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

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

cos (y) (e^{x} - 2)

Solution steps

\frac{\partial}{\partial x} ((e^{x} - 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} (e^{x} - 2 x)

Apply the Sum/Difference Rule:

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

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

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

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

= cos (y) (e^{x} - 2)

\frac{d y}{d x} + a y - b = 0

f (x) = 5 x^{2} - 4 x - 2, at x = 1

t y^{^{'}} + 10 y = \frac{1}{t}

\int 7 \frac{cos ( x ) + sin ( 2 x )}{sin ( x )} d x

1 - e^{- 2 x}

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