What is the d/(dt)(1+y/t) ?

The d/(dt)(1+y/t) is (t\frac{dy)/(dt)-y}{t^2}

What is the first d/(dt)(1+y/t) ?

The first d/(dt)(1+y/t) is (t\frac{dy)/(dt)-y}{t^2}

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Popular >

\frac{d}{d t} (1 + \frac{y}{t})

\frac{t \frac{d y}{d t} - y}{t ^{2}}

Solution steps

\frac{d}{d t} (1 + \frac{y}{t})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (1) + \frac{d}{d t} (\frac{y}{t})

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

\frac{d}{d t} (\frac{y}{t}) = \frac{t \frac{d y}{d t} - y}{t ^{2}}

= 0 + \frac{t \frac{d y}{d t} - y}{t ^{2}}

Simplify

= \frac{t \frac{d y}{d t} - y}{t ^{2}}

\int (x + 3) 5 - x d x

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\int \frac{162}{x ^{2} 81 - x ^{2}} d x

\int (cos (8 x)) d x

\frac{d}{d x} (\frac{x}{x ^{2} + 8})

What is the d/(dt)(1+y/t) ?
The d/(dt)(1+y/t) is (t\frac{dy)/(dt)-y}{t^2}
What is the first d/(dt)(1+y/t) ?
The first d/(dt)(1+y/t) is (t\frac{dy)/(dt)-y}{t^2}