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

The derivative of f(t)=(6t)/(6+sqrt(t)) is (3(sqrt(t)+12))/((6+sqrt(t))^2)

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

The first derivative of f(t)=(6t)/(6+sqrt(t)) is (3(sqrt(t)+12))/((6+sqrt(t))^2)

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derivative of

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

\frac{3 ( t + 12 )}{( 6 + t ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d t} (6 + t) = \frac{1}{2 t}

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

Simplify

6 \cdot \frac{1 \cdot ( 6 + t ) - \frac{1}{2 t} t}{( 6 + t ) ^{2}} : \frac{3 ( t + 12 )}{( 6 + t ) ^{2}}

= \frac{3 ( t + 12 )}{( 6 + t ) ^{2}}

x \to 0 + lim (x^{x^{\frac{1}{2}}})

\frac{d}{d y} (\frac{x}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

\int \frac{1}{x x ^{2} - π} d x

\int_{0}^{1} e^{x} - x d x

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What is the derivative of f(t)=(6t)/(6+sqrt(t)) ?
The derivative of f(t)=(6t)/(6+sqrt(t)) is (3(sqrt(t)+12))/((6+sqrt(t))^2)
What is the first derivative of f(t)=(6t)/(6+sqrt(t)) ?
The first derivative of f(t)=(6t)/(6+sqrt(t)) is (3(sqrt(t)+12))/((6+sqrt(t))^2)