What is the derivative of (1-6t)/(3+t) ?

The derivative of (1-6t)/(3+t) is -(19)/((3+t)^2)

What is the first derivative of (1-6t)/(3+t) ?

The first derivative of (1-6t)/(3+t) is -(19)/((3+t)^2)

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derivative of

\frac{1 - 6 t}{3 + t}

- \frac{19}{( 3 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1 - 6 t}{3 + 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 ) ( 3 + t ) - \frac{d}{d t} ( 3 + t ) ( 1 - 6 t )}{( 3 + t ) ^{2}}

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

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

= \frac{( - 6 ) ( 3 + t ) - 1 \cdot ( 1 - 6 t )}{( 3 + t ) ^{2}}

Simplify

\frac{( - 6 ) ( 3 + t ) - 1 \cdot ( 1 - 6 t )}{( 3 + t ) ^{2}} : - \frac{19}{( 3 + t ) ^{2}}

= - \frac{19}{( 3 + t ) ^{2}}

x \to 0 + lim (\frac{1}{x} + ln (x))

(x^{2} + 2 y^{2}) d x = - x y d y

\int x ln (2 x) d x

\frac{sec ( x )}{x}

\frac{d y}{d x} = e^{x + y}

What is the derivative of (1-6t)/(3+t) ?
The derivative of (1-6t)/(3+t) is -(19)/((3+t)^2)
What is the first derivative of (1-6t)/(3+t) ?
The first derivative of (1-6t)/(3+t) is -(19)/((3+t)^2)