What is (\partial}{\partial t}(\frac{2r+s)/t) ?

The answer to (\partial}{\partial t}(\frac{2r+s)/t) is -(2r+s)/(t^2)

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(\partial}{\partial t}(\frac{2r+s)/t)

\frac{\partial}{\partial t} (\frac{2 r + s}{t})

- \frac{2 r + s}{t ^{2}}

Solution steps

\frac{\partial}{\partial t} (\frac{2 r + s}{t})

Treat

r, s

as constants

Take the constant out:

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

= (2 r + s) \frac{\partial}{\partial t} (\frac{1}{t})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= (2 r + s) \frac{\partial}{\partial t} (t^{- 1})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= (2 r + s) (- 1 \cdot t^{- 1 - 1})

Simplify

(2 r + s) (- 1 \cdot t^{- 1 - 1}) : - \frac{2 r + s}{t ^{2}}

= - \frac{2 r + s}{t ^{2}}

\frac{d}{d x} (- 12 sin (6 x))

y = (3 x + 4)^{2}

x \to 0 lim (\frac{x ^{2} + 5}{3 ^{x}})

n = 1 \sum \infty (- 1)^{n} 2^{- n}

y = sin (2 x), at x = \frac{π}{4}

What is (\partial}{\partial t}(\frac{2r+s)/t) ?
The answer to (\partial}{\partial t}(\frac{2r+s)/t) is -(2r+s)/(t^2)