What is the derivative of (1-3t)/(5+t) ?

The derivative of (1-3t)/(5+t) is -(16)/((5+t)^2)

What is the first derivative of (1-3t)/(5+t) ?

The first derivative of (1-3t)/(5+t) is -(16)/((5+t)^2)

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derivative of

\frac{1 - 3 t}{5 + t}

- \frac{16}{( 5 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1 - 3 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 - 3 t ) ( 5 + t ) - \frac{d}{d t} ( 5 + t ) ( 1 - 3 t )}{( 5 + t ) ^{2}}

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

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

= \frac{( - 3 ) ( 5 + t ) - 1 \cdot ( 1 - 3 t )}{( 5 + t ) ^{2}}

Simplify

\frac{( - 3 ) ( 5 + t ) - 1 \cdot ( 1 - 3 t )}{( 5 + t ) ^{2}} : - \frac{16}{( 5 + t ) ^{2}}

= - \frac{16}{( 5 + t ) ^{2}}

y^{^{''}} - 10 y^{^{'}} + 24 y = 0

(r (θ) cos (θ))^{^{'}}

\int \frac{1}{2 u} d u

f (x) = e x

\int ln (1 + x^{2}) d x

What is the derivative of (1-3t)/(5+t) ?
The derivative of (1-3t)/(5+t) is -(16)/((5+t)^2)
What is the first derivative of (1-3t)/(5+t) ?
The first derivative of (1-3t)/(5+t) is -(16)/((5+t)^2)