What is the derivative of (9cos(x))/(1+sin(x)) ?

The derivative of (9cos(x))/(1+sin(x)) is (9(-sin(x)(1+sin(x))-cos^2(x)))/((1+sin(x))^2)

What is the first derivative of (9cos(x))/(1+sin(x)) ?

The first derivative of (9cos(x))/(1+sin(x)) is (9(-sin(x)(1+sin(x))-cos^2(x)))/((1+sin(x))^2)

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derivative of

\frac{9 cos ( x )}{1 + sin ( x )}

\frac{9 ( - sin ( x ) ( 1 + sin ( x ) ) - cos ^{2} ( x ) )}{( 1 + sin ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 cos ( x )}{1 + sin ( x )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 9 \frac{d}{d x} (\frac{cos ( x )}{1 + sin ( x )})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d}{d x} ( cos ( x ) ) ( 1 + sin ( x ) ) - \frac{d}{d x} ( 1 + sin ( x ) ) cos ( x )}{( 1 + sin ( x ) ) ^{2}}

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

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

= 9 \cdot \frac{( - sin ( x ) ) ( 1 + sin ( x ) ) - cos ( x ) cos ( x )}{( 1 + sin ( x ) ) ^{2}}

Simplify

9 \cdot \frac{( - sin ( x ) ) ( 1 + sin ( x ) ) - cos ( x ) cos ( x )}{( 1 + sin ( x ) ) ^{2}} : \frac{9 ( - sin ( x ) ( 1 + sin ( x ) ) - cos ^{2} ( x ) )}{( 1 + sin ( x ) ) ^{2}}

= \frac{9 ( - sin ( x ) ( 1 + sin ( x ) ) - cos ^{2} ( x ) )}{( 1 + sin ( x ) ) ^{2}}

\frac{1}{s - 3} - \frac{16}{s ^{2} + 9}

x \to 6 lim (ln (x - 6))

y = (3 x^{2} + 4) (3 + x^{3})^{5}

17 x, x^{2}

What is the derivative of (9cos(x))/(1+sin(x)) ?
The derivative of (9cos(x))/(1+sin(x)) is (9(-sin(x)(1+sin(x))-cos^2(x)))/((1+sin(x))^2)
What is the first derivative of (9cos(x))/(1+sin(x)) ?
The first derivative of (9cos(x))/(1+sin(x)) is (9(-sin(x)(1+sin(x))-cos^2(x)))/((1+sin(x))^2)