What is (\partial)/(\partial x)(ycos(x+y)) ?

The answer to (\partial)/(\partial x)(ycos(x+y)) is -ysin(x+y)

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(\partial)/(\partial x)(ycos(x+y))

\frac{\partial}{\partial x} (y cos (x + y))

- y sin (x + y)

Solution steps

\frac{\partial}{\partial x} (y cos (x + y))

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= y (- sin (x + y) \cdot 1)

Simplify

= - y sin (x + y)

\int 15 x \cdot 1 - x d x

\int \frac{cos ( 4 x ) cos ( 4 x )}{4} d x

\frac{4 - sec ( x )}{tan ( x )}

\frac{d}{d s} (\frac{s}{s ^{2} + 2})

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

What is (\partial)/(\partial x)(ycos(x+y)) ?
The answer to (\partial)/(\partial x)(ycos(x+y)) is -ysin(x+y)