What is the derivative of (sin(x)/(1+sin(x))) ?

The derivative of (sin(x)/(1+sin(x))) is (cos(x))/((1+sin(x))^2)

What is the first derivative of (sin(x)/(1+sin(x))) ?

The first derivative of (sin(x)/(1+sin(x))) is (cos(x))/((1+sin(x))^2)

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\frac{d}{d x} (\frac{sin ( x )}{1 + sin ( x )})

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

Solution steps

\frac{d}{d x} (\frac{sin ( x )}{1 + sin ( x )})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (1 + sin (x)) = cos (x)

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

Expand

cos (x) (1 + sin (x)) - cos (x) sin (x) : cos (x)

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

\int \frac{x ln ( x )}{x ^{2} - 1} d x

\int_{1}^{\infty} (\frac{1}{x ^{3}}) d x

\frac{d y}{d x} x + y = - 2 x^{6} y^{4}

4 \frac{d ^{2} x}{d t ^{2}} - 1 \frac{d x}{d t} - 3 x = 0, x (0) = 1

arctan (- 6 x)

What is the derivative of (sin(x)/(1+sin(x))) ?
The derivative of (sin(x)/(1+sin(x))) is (cos(x))/((1+sin(x))^2)
What is the first derivative of (sin(x)/(1+sin(x))) ?
The first derivative of (sin(x)/(1+sin(x))) is (cos(x))/((1+sin(x))^2)