What is the derivative of y=-(5t)/((t+8)^2) ?

The derivative of y=-(5t)/((t+8)^2) is -(5(8-t))/((t+8)^3)

What is the first derivative of y=-(5t)/((t+8)^2) ?

The first derivative of y=-(5t)/((t+8)^2) is -(5(8-t))/((t+8)^3)

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derivative of

y = - \frac{5 t}{( t + 8 ) ^{2}}

- \frac{5 ( 8 - t )}{( t + 8 ) ^{3}}

Solution steps

\frac{d}{d t} (- \frac{5 t}{( t + 8 ) ^{2}})

Take the constant out:

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

= - 5 \frac{d}{d t} (\frac{t}{( t + 8 ) ^{2}})

Apply the Quotient Rule:

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

= - 5 \cdot \frac{\frac{d t}{d t} ( t + 8 ) ^{2} - \frac{d}{d t} ( ( t + 8 ) ^{2} ) t}{( ( t + 8 ) ^{2} ) ^{2}}

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

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

= - 5 \cdot \frac{1 \cdot ( t + 8 ) ^{2} - 2 ( t + 8 ) t}{( ( t + 8 ) ^{2} ) ^{2}}

Simplify

- 5 \cdot \frac{1 \cdot ( t + 8 ) ^{2} - 2 ( t + 8 ) t}{( ( t + 8 ) ^{2} ) ^{2}} : - \frac{5 ( 8 - t )}{( t + 8 ) ^{3}}

= - \frac{5 ( 8 - t )}{( t + 8 ) ^{3}}

\int (x)^{3} arccos (x) d x

\int \frac{1}{1 + sin ( 2 x )} d x

\frac{2}{1 + x}

f (x) = \frac{x ^{3} + 2}{3 x}

f (x) = \frac{8}{x}, at x = 36

What is the derivative of y=-(5t)/((t+8)^2) ?
The derivative of y=-(5t)/((t+8)^2) is -(5(8-t))/((t+8)^3)
What is the first derivative of y=-(5t)/((t+8)^2) ?
The first derivative of y=-(5t)/((t+8)^2) is -(5(8-t))/((t+8)^3)