What is the derivative of G(t)=(1-6t)/(2+t) ?

The derivative of G(t)=(1-6t)/(2+t) is -(13)/((2+t)^2)

What is the first derivative of G(t)=(1-6t)/(2+t) ?

The first derivative of G(t)=(1-6t)/(2+t) is -(13)/((2+t)^2)

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derivative of

G (t) = \frac{1 - 6 t}{2 + t}

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

Solution steps

\frac{d}{d t} (\frac{1 - 6 t}{2 + t})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 1 - 6 t ) ( 2 + t ) - \frac{d}{d t} ( 2 + t ) ( 1 - 6 t )}{( 2 + t ) ^{2}}

\frac{d}{d t} (1 - 6 t) = - 6

\frac{d}{d t} (2 + t) = 1

= \frac{( - 6 ) ( 2 + t ) - 1 \cdot ( 1 - 6 t )}{( 2 + t ) ^{2}}

Simplify

\frac{( - 6 ) ( 2 + t ) - 1 \cdot ( 1 - 6 t )}{( 2 + t ) ^{2}} : - \frac{13}{( 2 + t ) ^{2}}

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

n = 1 \sum \infty (- \frac{3}{4})^{n}

x \to 0 - lim (cos (x^{2}))

\int_{0}^{3} \frac{1}{4 - t} d t

\frac{\partial}{\partial x} (- z^{2})

f (x) = 5 x - 1

What is the derivative of G(t)=(1-6t)/(2+t) ?
The derivative of G(t)=(1-6t)/(2+t) is -(13)/((2+t)^2)
What is the first derivative of G(t)=(1-6t)/(2+t) ?
The first derivative of G(t)=(1-6t)/(2+t) is -(13)/((2+t)^2)