What is the d/(dt)(x/t) ?

The d/(dt)(x/t) is (t\frac{dx)/(dt)-x}{t^2}

What is the first d/(dt)(x/t) ?

The first d/(dt)(x/t) is (t\frac{dx)/(dt)-x}{t^2}

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\frac{d}{d t} (\frac{x}{t})

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

Solution steps

\frac{d}{d t} (\frac{x}{t})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d t}{d t} = 1

= \frac{\frac{d x}{d t} t - 1 \cdot x}{t ^{2}}

Simplify

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

\int_{- \infty}^{- 1} 0 d x

\int \frac{18 x - 6}{3 x ^{2} - 2 x - 7} d x

\frac{10}{5 s + 1}

\int 4 sin (x) + \frac{2 x ^{5} - x}{x} d x

f (x) = - 3 x^{4} + 7 x^{2} - 4, at x = 1

What is the d/(dt)(x/t) ?
The d/(dt)(x/t) is (t\frac{dx)/(dt)-x}{t^2}
What is the first d/(dt)(x/t) ?
The first d/(dt)(x/t) is (t\frac{dx)/(dt)-x}{t^2}