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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d y} (y^{2})

\int \frac{1}{( x ln ( x ) )} d x

\int \frac{8 + 3 x}{x ( x ^{2} + 4 )} d x

x^{3} - 15 x^{2} + 72 x

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