What is the d/(dt)((5t)/(1+2t^2)) ?

The d/(dt)((5t)/(1+2t^2)) is (5(1-2t^2))/((1+2t^2)^2)

What is the first d/(dt)((5t)/(1+2t^2)) ?

The first d/(dt)((5t)/(1+2t^2)) is (5(1-2t^2))/((1+2t^2)^2)

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\frac{d}{d t} (\frac{5 t}{1 + 2 t ^{2}})

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

Solution steps

\frac{d}{d t} (\frac{5 t}{1 + 2 t ^{2}})

Take the constant out:

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

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

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

\frac{d}{d t} (1 + 2 t^{2}) = 4 t

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

Simplify

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

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

\frac{d y}{d t} = \frac{1}{( y + 1 ) ( t - 2 )}, y (0) = 0

\int cos (5 x) cos (10 x) d x

\frac{e ^{- s}}{( ( s - 3 ) ( s + 1 ) )}

\frac{d}{d x} (4 x^{\frac{- 1}{8}})

\frac{\partial}{\partial x} (5 x^{0.75} y^{0.25})

What is the d/(dt)((5t)/(1+2t^2)) ?
The d/(dt)((5t)/(1+2t^2)) is (5(1-2t^2))/((1+2t^2)^2)
What is the first d/(dt)((5t)/(1+2t^2)) ?
The first d/(dt)((5t)/(1+2t^2)) is (5(1-2t^2))/((1+2t^2)^2)