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

The d/(dt)((cos(t))/(sqrt(1+t^2))) is (-sin(t)(1+t^2)-tcos(t))/((1+t^2)sqrt(1+t^2))

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

The first d/(dt)((cos(t))/(sqrt(1+t^2))) is (-sin(t)(1+t^2)-tcos(t))/((1+t^2)sqrt(1+t^2))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( - sin ( t ) ) 1 + t ^{2} - \frac{t}{1 + t ^{2}} cos ( t )}{( 1 + t ^{2} ) ^{2}} : \frac{- sin ( t ) ( 1 + t ^{2} ) - t cos ( t )}{( 1 + t ^{2} ) 1 + t ^{2}}

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

\int \frac{x ^{5}}{a - b x ^{3}} d x

\frac{d}{d x} (\frac{x ^{5}}{4 - x ^{4}})

\frac{d}{d x} ((5 x)^{9 x})

\frac{d}{d x} (x^{4} + 3 x^{2})

f (x) = x^{4} + 3 x^{3}, at x = 1

What is the d/(dt)((cos(t))/(sqrt(1+t^2))) ?
The d/(dt)((cos(t))/(sqrt(1+t^2))) is (-sin(t)(1+t^2)-tcos(t))/((1+t^2)sqrt(1+t^2))
What is the first d/(dt)((cos(t))/(sqrt(1+t^2))) ?
The first d/(dt)((cos(t))/(sqrt(1+t^2))) is (-sin(t)(1+t^2)-tcos(t))/((1+t^2)sqrt(1+t^2))