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

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

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

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

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derivative of

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

- \frac{21}{( 5 + t ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - \frac{21}{( 5 + t ) ^{2}}

\int \frac{3}{8 + 3 x} d x

\int 5 - 4 x - 4 x^{2} d x

\frac{d}{d x} (\frac{- cos ( 2 x )}{2})

x \to - 1 + lim (\frac{3}{x ^{2} - 1})

\int_{0}^{1} x (e^{- x^{2}}) d x

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