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

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

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

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

2 y + sin (x + y)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (e^{y}) = 0

= 2 y - (- sin (x + y)) - 0

Simplify

= 2 y + sin (x + y)

n = 5 \sum \infty ln (n) - ln (2 n)

\frac{d}{d x} (- (cos (x)) ln (sec (x) + tan (x)))

x \to π lim (2 cot (x))

\int e^{t^{3}} d t

\int r^{2} (r^{3} + 5)^{8} d r

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