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

The derivative of y=(sin(x))/(1+cos(x)) is 1/(1+cos(x))

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

The first derivative of y=(sin(x))/(1+cos(x)) is 1/(1+cos(x))

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derivative of

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

\frac{1}{1 + cos ( x )}

Solution steps

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

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

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

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

Simplify

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

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

\int_{0}^{4} ∣ 4 x - 1 ∣ d x

\int \frac{6 x ^{2}}{4 x - x ^{2}} d x

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

n = 1 \sum \infty 6^{n} x^{n} n!

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