What is the d/(dt)((t+pi)/t cos(t)) ?

The d/(dt)((t+pi)/t cos(t)) is -(picos(t)}{t^2}-\frac{sin(t)(t+pi))/t

What is the first d/(dt)((t+pi)/t cos(t)) ?

The first d/(dt)((t+pi)/t cos(t)) is -(picos(t)}{t^2}-\frac{sin(t)(t+pi))/t

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\frac{d}{d t} (\frac{t + π}{t} cos (t))

- \frac{π cos ( t )}{t ^{2}} - \frac{sin ( t ) ( t + π )}{t}

Solution steps

\frac{d}{d t} (\frac{t + π}{t} cos (t))

Apply the Product Rule:

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

f = \frac{t + π}{t}, g = cos (t)

= \frac{d}{d t} (\frac{t + π}{t}) cos (t) + \frac{d}{d t} (cos (t)) \frac{t + π}{t}

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

\frac{d}{d t} (cos (t)) = - sin (t)

= (- \frac{π}{t ^{2}}) cos (t) + (- sin (t)) \frac{t + π}{t}

Simplify

(- \frac{π}{t ^{2}}) cos (t) + (- sin (t)) \frac{t + π}{t} : - \frac{π cos ( t )}{t ^{2}} - \frac{sin ( t ) ( t + π )}{t}

= - \frac{π cos ( t )}{t ^{2}} - \frac{sin ( t ) ( t + π )}{t}

x \to 0 lim (\frac{1}{x ^{2}} - \frac{1}{x})

\int 2 t + 1 d t

\int sec^{6} (3 x) d x

y^{^{'}} - \frac{4 y}{t} = 5

\frac{6 x ^{2} + 1}{x ^{2} + 9}

What is the d/(dt)((t+pi)/t cos(t)) ?
The d/(dt)((t+pi)/t cos(t)) is -(picos(t)}{t^2}-\frac{sin(t)(t+pi))/t
What is the first d/(dt)((t+pi)/t cos(t)) ?
The first d/(dt)((t+pi)/t cos(t)) is -(picos(t)}{t^2}-\frac{sin(t)(t+pi))/t