What is the d/(dt)((t(y^2+y))^{-1}) ?

The d/(dt)((t(y^2+y))^{-1}) is -(y^2+y+t(2y\frac{dy)/(dt)+(dy)/(dt))}{t^2(y^2+y)^2}

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

The first d/(dt)((t(y^2+y))^{-1}) is -(y^2+y+t(2y\frac{dy)/(dt)+(dy)/(dt))}{t^2(y^2+y)^2}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

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

Simplify

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

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

\int \frac{9 e ^{9 t}}{3 + e ^{9 t}} d t

\frac{d}{d x} (1 - 5 x)

n = 1 \sum \infty \frac{3}{2 ^{n}}

\int_{- 1}^{1} 5 x d x

\frac{d}{d t} (- 2 e^{- 2 t})

What is the d/(dt)((t(y^2+y))^{-1}) ?
The d/(dt)((t(y^2+y))^{-1}) is -(y^2+y+t(2y\frac{dy)/(dt)+(dy)/(dt))}{t^2(y^2+y)^2}
What is the first d/(dt)((t(y^2+y))^{-1}) ?
The first d/(dt)((t(y^2+y))^{-1}) is -(y^2+y+t(2y\frac{dy)/(dt)+(dy)/(dt))}{t^2(y^2+y)^2}