What is the derivative of (x^2+x-6/(x^2-4x+3)) ?

The derivative of (x^2+x-6/(x^2-4x+3)) is (-5x^2+18x-21)/((x^2-4x+3)^2)

What is the first derivative of (x^2+x-6/(x^2-4x+3)) ?

The first derivative of (x^2+x-6/(x^2-4x+3)) is (-5x^2+18x-21)/((x^2-4x+3)^2)

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\frac{d}{d x} (\frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3})

\frac{- 5 x ^{2} + 18 x - 21}{( x ^{2} - 4 x + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + x - 6 ) ( x ^{2} - 4 x + 3 ) - \frac{d}{d x} ( x ^{2} - 4 x + 3 ) ( x ^{2} + x - 6 )}{( x ^{2} - 4 x + 3 ) ^{2}}

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

\frac{d}{d x} (x^{2} - 4 x + 3) = 2 x - 4

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

Expand

(2 x + 1) (x^{2} - 4 x + 3) - (2 x - 4) (x^{2} + x - 6) : - 5 x^{2} + 18 x - 21

= \frac{- 5 x ^{2} + 18 x - 21}{( x ^{2} - 4 x + 3 ) ^{2}}

\frac{d}{d x} (4 sec (x) tan (x))

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int x^{11} x^{6} + 4 d x

\int 7 v (v^{2} + 2)^{2} d v

\frac{d}{d θ} (3 + 2 sin (θ))

What is the derivative of (x^2+x-6/(x^2-4x+3)) ?
The derivative of (x^2+x-6/(x^2-4x+3)) is (-5x^2+18x-21)/((x^2-4x+3)^2)
What is the first derivative of (x^2+x-6/(x^2-4x+3)) ?
The first derivative of (x^2+x-6/(x^2-4x+3)) is (-5x^2+18x-21)/((x^2-4x+3)^2)