What is the derivative of (1-3t)/(6+t) ?

The derivative of (1-3t)/(6+t) is -(19)/((6+t)^2)

What is the first derivative of (1-3t)/(6+t) ?

The first derivative of (1-3t)/(6+t) is -(19)/((6+t)^2)

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derivative of

\frac{1 - 3 t}{6 + t}

- \frac{19}{( 6 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1 - 3 t}{6 + t})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 1 - 3 t ) ( 6 + t ) - \frac{d}{d t} ( 6 + t ) ( 1 - 3 t )}{( 6 + t ) ^{2}}

\frac{d}{d t} (1 - 3 t) = - 3

\frac{d}{d t} (6 + t) = 1

= \frac{( - 3 ) ( 6 + t ) - 1 \cdot ( 1 - 3 t )}{( 6 + t ) ^{2}}

Simplify

\frac{( - 3 ) ( 6 + t ) - 1 \cdot ( 1 - 3 t )}{( 6 + t ) ^{2}} : - \frac{19}{( 6 + t ) ^{2}}

= - \frac{19}{( 6 + t ) ^{2}}

x \to 0 lim (- sin (x))

\frac{d}{d x} (2 x^{2} - 16 x + 35)

\int e^{t a n (7 x)} sec^{2} (7 x) d x

y = 10 x

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

What is the derivative of (1-3t)/(6+t) ?
The derivative of (1-3t)/(6+t) is -(19)/((6+t)^2)
What is the first derivative of (1-3t)/(6+t) ?
The first derivative of (1-3t)/(6+t) is -(19)/((6+t)^2)