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

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

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

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

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derivative of

G (t) = \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}}

\frac{\partial}{\partial y} (- \frac{y}{( x ^{2} + y ^{2} )})

\int e^{2 t} d t

\int \frac{9 - x ^{2}}{2 x ^{2}} d x

(x^{2} - 2 x + 2) e^{\frac{5 x}{2}}

x^{2} + y^{2} = 26

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