What is the derivative of y=(6x)/(9-tan(x)) ?

The derivative of y=(6x)/(9-tan(x)) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)

What is the first derivative of y=(6x)/(9-tan(x)) ?

The first derivative of y=(6x)/(9-tan(x)) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{6 x}{9 - tan ( x )}

\frac{6 ( 9 - tan ( x ) + x sec ^{2} ( x ) )}{( 9 - tan ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{9 - tan ( x )})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{9 - tan ( x )})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( 9 - tan ( x ) ) - \frac{d}{d x} ( 9 - tan ( x ) ) x}{( 9 - tan ( x ) ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (9 - tan (x)) = - sec^{2} (x)

= 6 \cdot \frac{1 \cdot ( 9 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 9 - tan ( x ) ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 9 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 9 - tan ( x ) ) ^{2}} : \frac{6 ( 9 - tan ( x ) + x sec ^{2} ( x ) )}{( 9 - tan ( x ) ) ^{2}}

= \frac{6 ( 9 - tan ( x ) + x sec ^{2} ( x ) )}{( 9 - tan ( x ) ) ^{2}}

\int - x^{4} e^{x} d x

\frac{d}{d x} (x z (x))

\int (x^{1.9} + 9 x^{3.5}) d x

\frac{d}{d x} (\frac{x}{x ^{- 1} + 4})

\frac{d}{d x} (1 + 2 e^{x})

What is the derivative of y=(6x)/(9-tan(x)) ?
The derivative of y=(6x)/(9-tan(x)) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)
What is the first derivative of y=(6x)/(9-tan(x)) ?
The first derivative of y=(6x)/(9-tan(x)) is (6(9-tan(x)+xsec^2(x)))/((9-tan(x))^2)