What is the derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) ?

The derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) is (2(x^2-3x+2))/((2x-3)^2)

What is the first derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) ?

The first derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) is (2(x^2-3x+2))/((2x-3)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = \frac{2 x ^{2} - 6 x + 5}{2 ( 2 x - 3 )}

\frac{2 ( x ^{2} - 3 x + 2 )}{( 2 x - 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} - 6 x + 5}{2 ( 2 x - 3 )})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{2 x ^{2} - 6 x + 5}{2 x - 3})

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( 2 x ^{2} - 6 x + 5 ) ( 2 x - 3 ) - \frac{d}{d x} ( 2 x - 3 ) ( 2 x ^{2} - 6 x + 5 )}{( 2 x - 3 ) ^{2}}

\frac{d}{d x} (2 x^{2} - 6 x + 5) = 4 x - 6

\frac{d}{d x} (2 x - 3) = 2

= \frac{1}{2} \cdot \frac{( 4 x - 6 ) ( 2 x - 3 ) - 2 ( 2 x ^{2} - 6 x + 5 )}{( 2 x - 3 ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{( 4 x - 6 ) ( 2 x - 3 ) - 2 ( 2 x ^{2} - 6 x + 5 )}{( 2 x - 3 ) ^{2}} : \frac{2 ( x ^{2} - 3 x + 2 )}{( 2 x - 3 ) ^{2}}

= \frac{2 ( x ^{2} - 3 x + 2 )}{( 2 x - 3 ) ^{2}}

\frac{d y}{d x} = 4 x e^{3 y}

x y^{^{'}} + 2 y = 4 cos (x)

\frac{d y}{d x} = sec^{2} (x) (2 + y)^{2}

\frac{4 e ^{- 2 s}}{3 s ^{2} + 48}

(2 y^{2} x - 3) d x + (2 y x^{2} + 4) d y = 0

What is the derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) ?
The derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) is (2(x^2-3x+2))/((2x-3)^2)
What is the first derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) ?
The first derivative of f(x)=(2x^2-6x+5)/(2(2x-3)) is (2(x^2-3x+2))/((2x-3)^2)