What is (\partial)/(\partial y)(2sin(x)+sin(x+y)) ?

The answer to (\partial)/(\partial y)(2sin(x)+sin(x+y)) is cos(x+y)

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(\partial)/(\partial y)(2sin(x)+sin(x+y))

\frac{\partial}{\partial y} (2 sin (x) + sin (x + y))

cos (x + y)

Solution steps

\frac{\partial}{\partial y} (2 sin (x) + sin (x + y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + cos (x + y)

Simplify

= cos (x + y)

x \to 0 lim (\frac{( 0 )}{x})

\frac{d y}{d x} + 5 y = 6

\int \frac{1}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (- x^{3} + 5 a x^{2} - 3 a^{2} x - b)

\int \frac{1}{10} x d x

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