What is the derivative of (x^2+1/(x^2-4x+3)) ?

The derivative of (x^2+1/(x^2-4x+3)) is (-4x^2+4x+4)/((x^2-4x+3)^2)

What is the first derivative of (x^2+1/(x^2-4x+3)) ?

The first derivative of (x^2+1/(x^2-4x+3)) is (-4x^2+4x+4)/((x^2-4x+3)^2)

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

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

Solution steps

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

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

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

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

Expand

2 x (x^{2} - 4 x + 3) - (2 x - 4) (x^{2} + 1) : - 4 x^{2} + 4 x + 4

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

\frac{d}{d x} (cos (x) + sin (x) + x)

\int \frac{x}{25 + x ^{4}} d x

\frac{d}{d x} (e^{x^{3}} \cdot 3 x^{2})

\frac{d}{d x} (2 x^{3} e^{- 0.6 x})

\int \frac{v}{v ^{2} + 1} d v

What is the derivative of (x^2+1/(x^2-4x+3)) ?
The derivative of (x^2+1/(x^2-4x+3)) is (-4x^2+4x+4)/((x^2-4x+3)^2)
What is the first derivative of (x^2+1/(x^2-4x+3)) ?
The first derivative of (x^2+1/(x^2-4x+3)) is (-4x^2+4x+4)/((x^2-4x+3)^2)