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

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

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

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

sin (t)

Solution steps

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

Treat

x, t

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + sin (t)

Simplify

= sin (t)

\int_{0}^{π} sin^{2} (x co) s^{4} x d x

\int (sec^{2} (x) + 2) d x

x \to - 7 + lim (\frac{x - 4}{x + 7})

\frac{\partial}{\partial x} (7 + 8 x y)

x \to 0 lim (\frac{1}{2 x})

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