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

The answer to (\partial)/(\partial x)(e^x-2-cos(xy)) is ysin(xy)+e^x

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

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

y sin (x y) + e^{x}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (2) = 0

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

= e^{x} - 0 - (- sin (x y) y)

Simplify

= y sin (x y) + e^{x}

\frac{d}{d x} (\frac{3}{5 + x})

\int \frac{1}{x + 4} d x

y^{^{'}} = \frac{2 cos ( y )}{sin ( y )}

\frac{d}{d x} (\frac{4 x ^{2} - x}{x ^{2} + 2})

\frac{6 x ^{2} + 2 x + 2}{x}

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