What is the d/(dt)((2t)/(2+sqrt(t))) ?

The d/(dt)((2t)/(2+sqrt(t))) is (4+sqrt(t))/((2+sqrt(t))^2)

What is the first d/(dt)((2t)/(2+sqrt(t))) ?

The first d/(dt)((2t)/(2+sqrt(t))) is (4+sqrt(t))/((2+sqrt(t))^2)

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\frac{d}{d t} (\frac{2 t}{2 + t})

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

Solution steps

\frac{d}{d t} (\frac{2 t}{2 + t})

Take the constant out:

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

= 2 \frac{d}{d t} (\frac{t}{2 + t})

Apply the Quotient Rule:

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

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

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

\frac{d}{d t} (2 + t) = \frac{1}{2 t}

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

Simplify

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

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

e^{2 x} sin (3 x)

x y^{^{''}} = y^{^{'}} + (y^{^{'}})^{2}

\int \frac{9 + x}{x ^{2}} d x

f (x) = - 3 - 7 x^{2}, at x = 42

\frac{d}{d x} (e^{- 0.05 x})

What is the d/(dt)((2t)/(2+sqrt(t))) ?
The d/(dt)((2t)/(2+sqrt(t))) is (4+sqrt(t))/((2+sqrt(t))^2)
What is the first d/(dt)((2t)/(2+sqrt(t))) ?
The first d/(dt)((2t)/(2+sqrt(t))) is (4+sqrt(t))/((2+sqrt(t))^2)