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Popular Calculus >

derivative of G(t)=(1-3t)/(4+t)

Solution

derivative of

G (t) = \frac{1 - 3 t}{4 + t}

Solution

- \frac{13}{( 4 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1 - 3 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 - 3 t ) ( 4 + t ) - \frac{d}{d t} ( 4 + t ) ( 1 - 3 t )}{( 4 + t ) ^{2}}

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

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

= \frac{( - 3 ) ( 4 + t ) - 1 \cdot ( 1 - 3 t )}{( 4 + t ) ^{2}}

Simplify

\frac{( - 3 ) ( 4 + t ) - 1 \cdot ( 1 - 3 t )}{( 4 + t ) ^{2}} : - \frac{13}{( 4 + t ) ^{2}}

= - \frac{13}{( 4 + t ) ^{2}}

Graph

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Frequently Asked Questions (FAQ)

What is the derivative of G(t)=(1-3t)/(4+t) ?
The derivative of G(t)=(1-3t)/(4+t) is -(13)/((4+t)^2)
What is the first derivative of G(t)=(1-3t)/(4+t) ?
The first derivative of G(t)=(1-3t)/(4+t) is -(13)/((4+t)^2)