What is (\partial)/(\partial t)(cos(t)-sin(t)) ?

The answer to (\partial)/(\partial t)(cos(t)-sin(t)) is -sin(t)-cos(t)

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(\partial)/(\partial t)(cos(t)-sin(t))

\frac{\partial}{\partial t} (cos (t) - sin (t))

- sin (t) - cos (t)

Solution steps

\frac{\partial}{\partial t} (cos (t) - sin (t))

Apply the Sum/Difference Rule:

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

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

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

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

= - sin (t) - cos (t)

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

\frac{\partial}{\partial x} (\frac{x + y}{1 + x ^{2}})

\int \frac{e ^{4 x}}{1 + e ^{4 x}} d x

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

What is (\partial)/(\partial t)(cos(t)-sin(t)) ?
The answer to (\partial)/(\partial t)(cos(t)-sin(t)) is -sin(t)-cos(t)