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)

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derivative of

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

- \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}}

f (x) = \frac{1}{x ^{3} + 2 x ^{2} - 1}

\int (e^{- \frac{x}{4}} + e^{\frac{x}{2}}) d x

(x + 1) \frac{d y}{d x} = (x + 6)

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

\int \frac{1}{( 5 x + 2 ) ^{3}} d x

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)