What is the derivative of (sin(t))/(4+4cos(t)) ?

The derivative of (sin(t))/(4+4cos(t)) is (cos(t)(4+4cos(t))+4sin^2(t))/((4+4cos(t))^2)

What is the first derivative of (sin(t))/(4+4cos(t)) ?

The first derivative of (sin(t))/(4+4cos(t)) is (cos(t)(4+4cos(t))+4sin^2(t))/((4+4cos(t))^2)

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derivative of

\frac{sin ( t )}{4 + 4 cos ( t )}

\frac{cos ( t ) ( 4 + 4 cos ( t ) ) + 4 sin ^{2} ( t )}{( 4 + 4 cos ( t ) ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{sin ( t )}{4 + 4 cos ( t )})

Apply the Quotient Rule:

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

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

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

\frac{d}{d t} (4 + 4 cos (t)) = - 4 sin (t)

= \frac{cos ( t ) ( 4 + 4 cos ( t ) ) - ( - 4 sin ( t ) ) sin ( t )}{( 4 + 4 cos ( t ) ) ^{2}}

Simplify

\frac{cos ( t ) ( 4 + 4 cos ( t ) ) - ( - 4 sin ( t ) ) sin ( t )}{( 4 + 4 cos ( t ) ) ^{2}} : \frac{cos ( t ) ( 4 + 4 cos ( t ) ) + 4 sin ^{2} ( t )}{( 4 + 4 cos ( t ) ) ^{2}}

= \frac{cos ( t ) ( 4 + 4 cos ( t ) ) + 4 sin ^{2} ( t )}{( 4 + 4 cos ( t ) ) ^{2}}

\frac{\partial}{\partial x} (x^{2} + x y + y^{2} + y)

\int_{0}^{\frac{π}{2}} (sin (x))^{2} d x

e^{- 2 t} \cdot sin (t)

\frac{\partial}{\partial y} (y sin (z))

x \to 0 lim ((tan (5 x))^{x})

What is the derivative of (sin(t))/(4+4cos(t)) ?
The derivative of (sin(t))/(4+4cos(t)) is (cos(t)(4+4cos(t))+4sin^2(t))/((4+4cos(t))^2)
What is the first derivative of (sin(t))/(4+4cos(t)) ?
The first derivative of (sin(t))/(4+4cos(t)) is (cos(t)(4+4cos(t))+4sin^2(t))/((4+4cos(t))^2)