What is the derivative of (440t)/((t^2+6)^3) ?

The derivative of (440t)/((t^2+6)^3) is (440(-5t^2+6))/((t^2+6)^4)

What is the first derivative of (440t)/((t^2+6)^3) ?

The first derivative of (440t)/((t^2+6)^3) is (440(-5t^2+6))/((t^2+6)^4)

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derivative of

\frac{440 t}{( t ^{2} + 6 ) ^{3}}

\frac{440 ( - 5 t ^{2} + 6 )}{( t ^{2} + 6 ) ^{4}}

Solution steps

\frac{d}{d t} (\frac{440 t}{( t ^{2} + 6 ) ^{3}})

Take the constant out:

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

= 440 \frac{d}{d t} (\frac{t}{( t ^{2} + 6 ) ^{3}})

Apply the Quotient Rule:

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

= 440 \cdot \frac{\frac{d t}{d t} ( t ^{2} + 6 ) ^{3} - \frac{d}{d t} ( ( t ^{2} + 6 ) ^{3} ) t}{( ( t ^{2} + 6 ) ^{3} ) ^{2}}

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

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

= 440 \cdot \frac{1 \cdot ( t ^{2} + 6 ) ^{3} - 6 t ( t ^{2} + 6 ) ^{2} t}{( ( t ^{2} + 6 ) ^{3} ) ^{2}}

Simplify

440 \cdot \frac{1 \cdot ( t ^{2} + 6 ) ^{3} - 6 t ( t ^{2} + 6 ) ^{2} t}{( ( t ^{2} + 6 ) ^{3} ) ^{2}} : \frac{440 ( - 5 t ^{2} + 6 )}{( t ^{2} + 6 ) ^{4}}

= \frac{440 ( - 5 t ^{2} + 6 )}{( t ^{2} + 6 ) ^{4}}

f (x) = sin^{5} (3 x)

\frac{7 x}{x + 4} (3.3)

\frac{d}{d x} ((x^{3} + x)^{6})

ln (x + x^{2} - 10)

sin (9 x)

What is the derivative of (440t)/((t^2+6)^3) ?
The derivative of (440t)/((t^2+6)^3) is (440(-5t^2+6))/((t^2+6)^4)
What is the first derivative of (440t)/((t^2+6)^3) ?
The first derivative of (440t)/((t^2+6)^3) is (440(-5t^2+6))/((t^2+6)^4)