What is the derivative of g(x)=(1+6x)/(9-4x) ?

The derivative of g(x)=(1+6x)/(9-4x) is (58)/((9-4x)^2)

What is the first derivative of g(x)=(1+6x)/(9-4x) ?

The first derivative of g(x)=(1+6x)/(9-4x) is (58)/((9-4x)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

g (x) = \frac{1 + 6 x}{9 - 4 x}

\frac{58}{( 9 - 4 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1 + 6 x}{9 - 4 x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 1 + 6 x ) ( 9 - 4 x ) - \frac{d}{d x} ( 9 - 4 x ) ( 1 + 6 x )}{( 9 - 4 x ) ^{2}}

\frac{d}{d x} (1 + 6 x) = 6

\frac{d}{d x} (9 - 4 x) = - 4

= \frac{6 ( 9 - 4 x ) - ( - 4 ) ( 1 + 6 x )}{( 9 - 4 x ) ^{2}}

Simplify

\frac{6 ( 9 - 4 x ) - ( - 4 ) ( 1 + 6 x )}{( 9 - 4 x ) ^{2}} : \frac{58}{( 9 - 4 x ) ^{2}}

= \frac{58}{( 9 - 4 x ) ^{2}}

f^{^{'}} (x) = 3^{x}

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

4 x^{2} + 1

4 y^{^{''}} - 4 y^{^{'}} + 10 y = 0

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

What is the derivative of g(x)=(1+6x)/(9-4x) ?
The derivative of g(x)=(1+6x)/(9-4x) is (58)/((9-4x)^2)
What is the first derivative of g(x)=(1+6x)/(9-4x) ?
The first derivative of g(x)=(1+6x)/(9-4x) is (58)/((9-4x)^2)