What is the derivative of 1/(x-1+1/(4-x)) ?

The derivative of 1/(x-1+1/(4-x)) is (6x-15)/((x-1)^2(-x+4)^2)

What is the first derivative of 1/(x-1+1/(4-x)) ?

The first derivative of 1/(x-1+1/(4-x)) is (6x-15)/((x-1)^2(-x+4)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\frac{6 x - 15}{( x - 1 ) ^{2} ( - x + 4 ) ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

Simplify

- \frac{1}{( x - 1 ) ^{2}} + \frac{1}{( 4 - x ) ^{2}} : \frac{6 x - 15}{( x - 1 ) ^{2} ( - x + 4 ) ^{2}}

= \frac{6 x - 15}{( x - 1 ) ^{2} ( - x + 4 ) ^{2}}

\frac{d}{d x} (arcsin (x y))

f (x) = \frac{x e ^{x}}{x + e ^{x}}

\int \frac{4}{( x + 2 ) ( x - 2 )} d x

y^{^{'}} = \frac{2 cos ( 2 x )}{3 + 2 y}, y (0) = - 1

\int_{0}^{1} \frac{x - 1}{x ^{2} + 3 x + 2} d x

What is the derivative of 1/(x-1+1/(4-x)) ?
The derivative of 1/(x-1+1/(4-x)) is (6x-15)/((x-1)^2(-x+4)^2)
What is the first derivative of 1/(x-1+1/(4-x)) ?
The first derivative of 1/(x-1+1/(4-x)) is (6x-15)/((x-1)^2(-x+4)^2)