What is the derivative of G(t)=(4t)/(t+1) ?

The derivative of G(t)=(4t)/(t+1) is 4/((t+1)^2)

What is the first derivative of G(t)=(4t)/(t+1) ?

The first derivative of G(t)=(4t)/(t+1) is 4/((t+1)^2)

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derivative of

G (t) = \frac{4 t}{t + 1}

\frac{4}{( t + 1 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{4 t}{t + 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d t} (\frac{t}{t + 1})

Apply the Quotient Rule:

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

= 4 \cdot \frac{\frac{d t}{d t} ( t + 1 ) - \frac{d}{d t} ( t + 1 ) t}{( t + 1 ) ^{2}}

\frac{d t}{d t} = 1

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

= 4 \cdot \frac{1 \cdot ( t + 1 ) - 1 \cdot t}{( t + 1 ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( t + 1 ) - 1 \cdot t}{( t + 1 ) ^{2}} : \frac{4}{( t + 1 ) ^{2}}

= \frac{4}{( t + 1 ) ^{2}}

f (x) = x^{2} + 5 x, at x = 4

\int x e^{- x} - e^{- x} d x

\frac{\partial}{\partial x} (3 e^{3 x})

y^{^{'}} = y (1 - y), y (0) = 8

x cos (x) - sin (x)

What is the derivative of G(t)=(4t)/(t+1) ?
The derivative of G(t)=(4t)/(t+1) is 4/((t+1)^2)
What is the first derivative of G(t)=(4t)/(t+1) ?
The first derivative of G(t)=(4t)/(t+1) is 4/((t+1)^2)