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

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

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

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

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

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

Solution steps

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

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

\frac{d}{d x} (1 + tan (x)) = sec^{2} (x)

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

Simplify

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

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

x + \frac{3}{x}, at x = 2

f (x) = 4 - 2 x

x \to 2 lim (\frac{( x ^{6} - 64 )}{x - 2})

f (x) = 2 x

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

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