What is the derivative of (480t)/((t^2+5)^3) ?

The derivative of (480t)/((t^2+5)^3) is (2400(-t^2+1))/((t^2+5)^4)

What is the first derivative of (480t)/((t^2+5)^3) ?

The first derivative of (480t)/((t^2+5)^3) is (2400(-t^2+1))/((t^2+5)^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{480 t}{( t ^{2} + 5 ) ^{3}}

\frac{2400 ( - t ^{2} + 1 )}{( t ^{2} + 5 ) ^{4}}

Solution steps

\frac{d}{d t} (\frac{480 t}{( t ^{2} + 5 ) ^{3}})

Take the constant out:

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

= 480 \frac{d}{d t} (\frac{t}{( t ^{2} + 5 ) ^{3}})

Apply the Quotient Rule:

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

= 480 \cdot \frac{\frac{d t}{d t} ( t ^{2} + 5 ) ^{3} - \frac{d}{d t} ( ( t ^{2} + 5 ) ^{3} ) t}{( ( t ^{2} + 5 ) ^{3} ) ^{2}}

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

\frac{d}{d t} ((t^{2} + 5)^{3}) = 6 t (t^{2} + 5)^{2}

= 480 \cdot \frac{1 \cdot ( t ^{2} + 5 ) ^{3} - 6 t ( t ^{2} + 5 ) ^{2} t}{( ( t ^{2} + 5 ) ^{3} ) ^{2}}

Simplify

480 \cdot \frac{1 \cdot ( t ^{2} + 5 ) ^{3} - 6 t ( t ^{2} + 5 ) ^{2} t}{( ( t ^{2} + 5 ) ^{3} ) ^{2}} : \frac{2400 ( - t ^{2} + 1 )}{( t ^{2} + 5 ) ^{4}}

= \frac{2400 ( - t ^{2} + 1 )}{( t ^{2} + 5 ) ^{4}}

\int 5 e^{- 0.1 x} d x

x \to 1 lim (- 3 x + 5)

t \frac{d y}{d t} + (t + 1) y = 2 t e^{- t}

3 x^{2} + 4 x - 2

y = \frac{3}{x}, at x = \frac{1}{16}

What is the derivative of (480t)/((t^2+5)^3) ?
The derivative of (480t)/((t^2+5)^3) is (2400(-t^2+1))/((t^2+5)^4)
What is the first derivative of (480t)/((t^2+5)^3) ?
The first derivative of (480t)/((t^2+5)^3) is (2400(-t^2+1))/((t^2+5)^4)