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

The derivative of f(x)=(x^2-4x+3)/(x+1) is (x^2+2x-7)/((x+1)^2)

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

The first derivative of f(x)=(x^2-4x+3)/(x+1) is (x^2+2x-7)/((x+1)^2)

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derivative of

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

\frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 x - 4 ) ( x + 1 ) - 1 \cdot ( x ^{2} - 4 x + 3 )}{( x + 1 ) ^{2}} : \frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

= \frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

\int \frac{1}{1 - tan ( 6 x )} d x

x \to 1 lim (x^{3} - 2 x^{2} - 7 x - 6)

x \to \infty lim ((- 1)^{x})

\int \frac{1}{4 t t ^{2} - 16} d t

y^{^{'}} = 2 x^{2}

What is the derivative of f(x)=(x^2-4x+3)/(x+1) ?
The derivative of f(x)=(x^2-4x+3)/(x+1) is (x^2+2x-7)/((x+1)^2)
What is the first derivative of f(x)=(x^2-4x+3)/(x+1) ?
The first derivative of f(x)=(x^2-4x+3)/(x+1) is (x^2+2x-7)/((x+1)^2)