What is the derivative of y=(5(1-sin(x)))/(2cos(x)) ?

The derivative of y=(5(1-sin(x)))/(2cos(x)) is (5(-1+sin(x)))/(2cos^2(x))

What is the first derivative of y=(5(1-sin(x)))/(2cos(x)) ?

The first derivative of y=(5(1-sin(x)))/(2cos(x)) is (5(-1+sin(x)))/(2cos^2(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{5 ( 1 - sin ( x ) )}{2 cos ( x )}

\frac{5 ( - 1 + sin ( x ) )}{2 cos ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{5 ( 1 - sin ( x ) )}{2 cos ( x )})

Take the constant out:

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

= \frac{5}{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{5}{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{5}{2} \cdot \frac{( - cos ( x ) ) cos ( x ) - ( - sin ( x ) ) ( 1 - sin ( x ) )}{( cos ( x ) ) ^{2}}

Simplify

\frac{5}{2} \cdot \frac{( - cos ( x ) ) cos ( x ) - ( - sin ( x ) ) ( 1 - sin ( x ) )}{cos ^{2} ( x )} : \frac{5 ( - 1 + sin ( x ) )}{2 cos ^{2} ( x )}

= \frac{5 ( - 1 + sin ( x ) )}{2 cos ^{2} ( x )}

tan^{4} (x^{3})

f (x) = x^{2}, x = 5

f (x) = \frac{sin ( x )}{1 + 2 cos ( x )}

y^{^{'}} = y - 2

\frac{\partial}{\partial x} (x ln (z^{7} + x^{3}))

What is the derivative of y=(5(1-sin(x)))/(2cos(x)) ?
The derivative of y=(5(1-sin(x)))/(2cos(x)) is (5(-1+sin(x)))/(2cos^2(x))
What is the first derivative of y=(5(1-sin(x)))/(2cos(x)) ?
The first derivative of y=(5(1-sin(x)))/(2cos(x)) is (5(-1+sin(x)))/(2cos^2(x))