What is the derivative of (1-6t)/(5+t) ?

The derivative of (1-6t)/(5+t) is -(31)/((5+t)^2)

What is the first derivative of (1-6t)/(5+t) ?

The first derivative of (1-6t)/(5+t) is -(31)/((5+t)^2)

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derivative of

\frac{1 - 6 t}{5 + t}

- \frac{31}{( 5 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1 - 6 t}{5 + 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 ) ( 5 + t ) - \frac{d}{d t} ( 5 + t ) ( 1 - 6 t )}{( 5 + t ) ^{2}}

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

\frac{d}{d t} (5 + t) = 1

= \frac{( - 6 ) ( 5 + t ) - 1 \cdot ( 1 - 6 t )}{( 5 + t ) ^{2}}

Simplify

\frac{( - 6 ) ( 5 + t ) - 1 \cdot ( 1 - 6 t )}{( 5 + t ) ^{2}} : - \frac{31}{( 5 + t ) ^{2}}

= - \frac{31}{( 5 + t ) ^{2}}

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\frac{d}{d x} (1 - e^{3 x})

\int_{0}^{2} x^{2} (1 - 0.5 x) d x

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What is the derivative of (1-6t)/(5+t) ?
The derivative of (1-6t)/(5+t) is -(31)/((5+t)^2)
What is the first derivative of (1-6t)/(5+t) ?
The first derivative of (1-6t)/(5+t) is -(31)/((5+t)^2)