What is the derivative of y=(8t)/(8+sqrt(t)) ?

The derivative of y=(8t)/(8+sqrt(t)) is (4(sqrt(t)+16))/((8+sqrt(t))^2)

What is the first derivative of y=(8t)/(8+sqrt(t)) ?

The first derivative of y=(8t)/(8+sqrt(t)) is (4(sqrt(t)+16))/((8+sqrt(t))^2)

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derivative of

y = \frac{8 t}{8 + t}

\frac{4 ( t + 16 )}{( 8 + t ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{8 t}{8 + t})

Take the constant out:

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

= 8 \frac{d}{d t} (\frac{t}{8 + t})

Apply the Quotient Rule:

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

= 8 \cdot \frac{\frac{d t}{d t} ( 8 + t ) - \frac{d}{d t} ( 8 + t ) t}{( 8 + t ) ^{2}}

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

\frac{d}{d t} (8 + t) = \frac{1}{2 t}

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

Simplify

8 \cdot \frac{1 \cdot ( 8 + t ) - \frac{1}{2 t} t}{( 8 + t ) ^{2}} : \frac{4 ( t + 16 )}{( 8 + t ) ^{2}}

= \frac{4 ( t + 16 )}{( 8 + t ) ^{2}}

x \to 2 lim (2 x e^{x^{2}})

\int 2 x^{3} (1 - x^{4})^{- \frac{1}{4}} d x

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\frac{d}{d x} (\frac{b}{x})

What is the derivative of y=(8t)/(8+sqrt(t)) ?
The derivative of y=(8t)/(8+sqrt(t)) is (4(sqrt(t)+16))/((8+sqrt(t))^2)
What is the first derivative of y=(8t)/(8+sqrt(t)) ?
The first derivative of y=(8t)/(8+sqrt(t)) is (4(sqrt(t)+16))/((8+sqrt(t))^2)