What is (\partial)/(\partial s)(scos(t-pi/2)) ?

The answer to (\partial)/(\partial s)(scos(t-pi/2)) is sin(t)

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(\partial)/(\partial s)(scos(t-pi/2))

\frac{\partial}{\partial s} (s cos (t - \frac{π}{2}))

sin (t)

Solution steps

\frac{\partial}{\partial s} (s cos (t - \frac{π}{2}))

Simplify

s cos (t - \frac{π}{2}) : s sin (t)

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

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the common derivative:

\frac{\partial}{\partial s} (s) = 1

= sin (t) \cdot 1

Simplify

= sin (t)

\frac{d}{d x} (\frac{- 3}{x ^{2} + 9})

\int \frac{x ^{2}}{a - x ^{2}} d x

\frac{d}{d x} (csc (x) cot (x))

f (t) = 80 e^{- \frac{t}{10}} - 40 t

\frac{x}{x - 7}

What is (\partial)/(\partial s)(scos(t-pi/2)) ?
The answer to (\partial)/(\partial s)(scos(t-pi/2)) is sin(t)