What is (\partial)/(\partial y)(sin(y)+y,x) ?

The answer to (\partial)/(\partial y)(sin(y)+y,x) is cos(y)+(,x*1)

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(\partial)/(\partial y)(sin(y)+y,x)

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

cos (y) + (, x \cdot 1)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (y, x) = (, x \cdot 1)

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

\int_{0}^{2} t e^{t} d t

sin^{2} (x), sin^{3} (x), 0, π

\int 3 sin (x) + 2 - 3 e^{x} d x

f (x) = 4 x (1 - x), g (x) = 4 x - 4

\frac{d}{d x} (\frac{6 x ^{3} - 5 x ^{2} - 2 x}{4 x - 3 x ^{2}})

What is (\partial)/(\partial y)(sin(y)+y,x) ?
The answer to (\partial)/(\partial y)(sin(y)+y,x) is cos(y)+(,x*1)