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

The answer to (\partial)/(\partial x)((e^x-x)cos(y)) is cos(y)(e^x-1)

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

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

cos (y) (e^{x} - 1)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (x) = 1

= cos (y) (e^{x} - 1)

\frac{d}{d t} (- 32 t)

\int 25 - x^{2} - 3 d x

x \to 3 + lim (\frac{1}{x + 3})

(x + e^{y}) y^{^{'}} = x e^{- y} - 1

2^{x - 1}

What is (\partial)/(\partial x)((e^x-x)cos(y)) ?
The answer to (\partial)/(\partial x)((e^x-x)cos(y)) is cos(y)(e^x-1)