What is the derivative of (1-5t)/(4+t) ?

The derivative of (1-5t)/(4+t) is -(21)/((4+t)^2)

What is the first derivative of (1-5t)/(4+t) ?

The first derivative of (1-5t)/(4+t) is -(21)/((4+t)^2)

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derivative of

\frac{1 - 5 t}{4 + t}

- \frac{21}{( 4 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1 - 5 t}{4 + t})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 1 - 5 t ) ( 4 + t ) - \frac{d}{d t} ( 4 + t ) ( 1 - 5 t )}{( 4 + t ) ^{2}}

\frac{d}{d t} (1 - 5 t) = - 5

\frac{d}{d t} (4 + t) = 1

= \frac{( - 5 ) ( 4 + t ) - 1 \cdot ( 1 - 5 t )}{( 4 + t ) ^{2}}

Simplify

\frac{( - 5 ) ( 4 + t ) - 1 \cdot ( 1 - 5 t )}{( 4 + t ) ^{2}} : - \frac{21}{( 4 + t ) ^{2}}

= - \frac{21}{( 4 + t ) ^{2}}

\int sin (8 x) cos (8 x) d x

(x - 1)^{2}

\frac{d}{d x} (sec^{2} (x) d x)

(6 - 10 y + e^{- 5 x}) d x - 2 d y = 0

\frac{2 - s}{s ^{2} + 6 s + 34}

What is the derivative of (1-5t)/(4+t) ?
The derivative of (1-5t)/(4+t) is -(21)/((4+t)^2)
What is the first derivative of (1-5t)/(4+t) ?
The first derivative of (1-5t)/(4+t) is -(21)/((4+t)^2)