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

The d/(dt)((x)(2/(t^2+1)-1)) is (-t^4\frac{dx)/(dt)-4tx+(dx)/(dt)}{(t^2+1)^2}

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

The first d/(dt)((x)(2/(t^2+1)-1)) is (-t^4\frac{dx)/(dt)-4tx+(dx)/(dt)}{(t^2+1)^2}

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = \frac{2}{t ^{2} + 1} - 1

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

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

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

Simplify

\frac{d x}{d t} (\frac{2}{t ^{2} + 1} - 1) + (- \frac{4 t}{( t ^{2} + 1 ) ^{2}}) x : \frac{- t ^{4} \frac{d x}{d t} - 4 t x + \frac{d x}{d t}}{( t ^{2} + 1 ) ^{2}}

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

\frac{( 7 q ^{2} + 4 ) ^{2}}{49}

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

5 (x^{2} + y^{2}) e^{y^{2} - x^{2}}

\frac{a ^{2} + 4 a + 4}{( 3 a + 3 b ) ( a + b )}

\frac{3 x ^{2} + 1}{( x + 2 ) ^{3}}

What is the d/(dt)((x)(2/(t^2+1)-1)) ?
The d/(dt)((x)(2/(t^2+1)-1)) is (-t^4\frac{dx)/(dt)-4tx+(dx)/(dt)}{(t^2+1)^2}
What is the first d/(dt)((x)(2/(t^2+1)-1)) ?
The first d/(dt)((x)(2/(t^2+1)-1)) is (-t^4\frac{dx)/(dt)-4tx+(dx)/(dt)}{(t^2+1)^2}