What is the d/(db)(((1+sin(b)))/((1-sin(b)))) ?

The d/(db)(((1+sin(b)))/((1-sin(b)))) is (2cos(b))/((1-sin(b))^2)

What is the first d/(db)(((1+sin(b)))/((1-sin(b)))) ?

The first d/(db)(((1+sin(b)))/((1-sin(b)))) is (2cos(b))/((1-sin(b))^2)

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\frac{d}{d b} (\frac{( 1 + sin ( b ) )}{( 1 - sin ( b ) )})

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

Solution steps

\frac{d}{d b} (\frac{( 1 + sin ( b ) )}{( 1 - sin ( b ) )})

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y = \frac{x}{( x + 2 )}

- 2 y^{2}

ln (x^{4} + 3)

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

64 (2 + x)^{1}

What is the d/(db)(((1+sin(b)))/((1-sin(b)))) ?
The d/(db)(((1+sin(b)))/((1-sin(b)))) is (2cos(b))/((1-sin(b))^2)
What is the first d/(db)(((1+sin(b)))/((1-sin(b)))) ?
The first d/(db)(((1+sin(b)))/((1-sin(b)))) is (2cos(b))/((1-sin(b))^2)