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

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

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

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

sin (x + y) + y cos (x + y)

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = sin (x + y)

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

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

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

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

Simplify

= sin (x + y) + y cos (x + y)

\frac{d}{d x} (e^{6 x} ln (x^{2} - 4))

\int_{- 1}^{2} (x - 6 ∣ x ∣) d x

\int \frac{1}{- x ^{2} + 6 x - 5} d x

4 x^{3} y + x^{4} y^{^{'}} = sin^{3} (\frac{x}{9})

ln (x + 1), 0

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