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

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

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

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

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

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

Solution steps

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

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

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

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

Simplify

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

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

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

e^{- 2 x} sin (x), 0

\int \frac{5 x ^{2} + 5 x + 3}{x ( x + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}})

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = - 3

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