What is the derivative of f(t)=(6t+38)/(t+9) ?

The derivative of f(t)=(6t+38)/(t+9) is (16)/((t+9)^2)

What is the first derivative of f(t)=(6t+38)/(t+9) ?

The first derivative of f(t)=(6t+38)/(t+9) is (16)/((t+9)^2)

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derivative of

f (t) = \frac{6 t + 38}{t + 9}

\frac{16}{( t + 9 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{6 t + 38}{t + 9})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 6 t + 38 ) ( t + 9 ) - \frac{d}{d t} ( t + 9 ) ( 6 t + 38 )}{( t + 9 ) ^{2}}

\frac{d}{d t} (6 t + 38) = 6

\frac{d}{d t} (t + 9) = 1

= \frac{6 ( t + 9 ) - 1 \cdot ( 6 t + 38 )}{( t + 9 ) ^{2}}

Simplify

\frac{6 ( t + 9 ) - 1 \cdot ( 6 t + 38 )}{( t + 9 ) ^{2}} : \frac{16}{( t + 9 ) ^{2}}

= \frac{16}{( t + 9 ) ^{2}}

\int_{1}^{6} \frac{x ^{2} + 1}{7 x - x ^{2}} d x

\frac{d}{d x} (π x)

\frac{d y}{d x} + \frac{4}{x} y = - 3 x

\frac{1}{( s ^{2} + 2 s + 1 )}

\frac{d}{d x} (\frac{e ^{x}}{( 1 - e ^{x} ) ^{2}})

What is the derivative of f(t)=(6t+38)/(t+9) ?
The derivative of f(t)=(6t+38)/(t+9) is (16)/((t+9)^2)
What is the first derivative of f(t)=(6t+38)/(t+9) ?
The first derivative of f(t)=(6t+38)/(t+9) is (16)/((t+9)^2)