What is the derivative of f(t)=(4t)/(4+sqrt(t)) ?

The derivative of f(t)=(4t)/(4+sqrt(t)) is (2(sqrt(t)+8))/((4+sqrt(t))^2)

What is the first derivative of f(t)=(4t)/(4+sqrt(t)) ?

The first derivative of f(t)=(4t)/(4+sqrt(t)) is (2(sqrt(t)+8))/((4+sqrt(t))^2)

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derivative of

f (t) = \frac{4 t}{4 + t}

\frac{2 ( t + 8 )}{( 4 + t ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

4 \cdot \frac{1 \cdot ( 4 + t ) - \frac{1}{2 t} t}{( 4 + t ) ^{2}} : \frac{2 ( t + 8 )}{( 4 + t ) ^{2}}

= \frac{2 ( t + 8 )}{( 4 + t ) ^{2}}

\int \frac{9}{1 - x ^{2}} d x

\frac{d}{d x} (ln (5 x - 4))

\int \frac{1}{x ^{2} - 6 x + 34} d x

y^{^{''}} + y = - sin (3 x)

\frac{d}{d x} (2 sin (π x))

What is the derivative of f(t)=(4t)/(4+sqrt(t)) ?
The derivative of f(t)=(4t)/(4+sqrt(t)) is (2(sqrt(t)+8))/((4+sqrt(t))^2)
What is the first derivative of f(t)=(4t)/(4+sqrt(t)) ?
The first derivative of f(t)=(4t)/(4+sqrt(t)) is (2(sqrt(t)+8))/((4+sqrt(t))^2)