What is (\partial}{\partial s}(\frac{(s-3r))/t) ?

The answer to (\partial}{\partial s}(\frac{(s-3r))/t) is 1/t

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(\partial}{\partial s}(\frac{(s-3r))/t)

\frac{\partial}{\partial s} (\frac{( s - 3 r )}{t})

\frac{1}{t}

Solution steps

\frac{\partial}{\partial s} (\frac{( s - 3 r )}{t})

Treat

r, t

as constants

Take the constant out:

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

= \frac{1}{t} \frac{\partial}{\partial s} (s - 3 r)

Apply the Sum/Difference Rule:

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

= \frac{1}{t} (\frac{\partial}{\partial s} (s) - \frac{\partial}{\partial s} (3 r))

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

\frac{\partial}{\partial s} (3 r) = 0

= \frac{1}{t} (1 - 0)

Simplify

= \frac{1}{t}

\frac{d}{d x} (sin^{2} (x) + cos (x))

\int x^{2} 13 x - 1 d x

\int - \frac{1}{2 y} d y

\int - 4 sin^{2} (x) d x

y = \frac{1}{x ^{3}}, (- 2, - \frac{1}{8})

What is (\partial}{\partial s}(\frac{(s-3r))/t) ?
The answer to (\partial}{\partial s}(\frac{(s-3r))/t) is 1/t