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

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

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

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

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derivative of

y = \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{\partial}{\partial x} (ln (x y))

x \to 0 lim (e^{- \frac{1}{x ^{2}}})

\frac{6}{2 - 4 x}

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

7 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

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