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

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

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

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

y cos (t)

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + y cos (t)

Simplify

= y cos (t)

\int t^{2} - 9 d t

\int (3 x^{2} + 4 x + 2) d x

\frac{d}{d x} (\frac{48}{( 1 - 2 x ) ^{4}})

\frac{d}{d x} (ln (x + 2 + x^{2}))

x y^{2} y^{^{'}} = x^{3} + y^{3}

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