What is the derivative of f(x)=((x+9))/(x^2-7x+1) ?

The derivative of f(x)=((x+9))/(x^2-7x+1) is (-x^2-18x+64)/((x^2-7x+1)^2)

What is the first derivative of f(x)=((x+9))/(x^2-7x+1) ?

The first derivative of f(x)=((x+9))/(x^2-7x+1) is (-x^2-18x+64)/((x^2-7x+1)^2)

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derivative of

f (x) = \frac{( x + 9 )}{x ^{2} - 7 x + 1}

\frac{- x ^{2} - 18 x + 64}{( x ^{2} - 7 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x + 9}{x ^{2} - 7 x + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x + 9 ) ( x ^{2} - 7 x + 1 ) - \frac{d}{d x} ( x ^{2} - 7 x + 1 ) ( x + 9 )}{( x ^{2} - 7 x + 1 ) ^{2}}

\frac{d}{d x} (x + 9) = 1

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

= \frac{1 \cdot ( x ^{2} - 7 x + 1 ) - ( 2 x - 7 ) ( x + 9 )}{( x ^{2} - 7 x + 1 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{2} - 7 x + 1 ) - ( 2 x - 7 ) ( x + 9 )}{( x ^{2} - 7 x + 1 ) ^{2}} : \frac{- x ^{2} - 18 x + 64}{( x ^{2} - 7 x + 1 ) ^{2}}

= \frac{- x ^{2} - 18 x + 64}{( x ^{2} - 7 x + 1 ) ^{2}}

g (x) = \frac{7}{x}

\frac{s}{( s + 1 ) ^{2}}

\int \frac{4 x ^{5}}{1 - 12 x ^{6}} d x

12 x^{3} - 18 x^{2}

What is the derivative of f(x)=((x+9))/(x^2-7x+1) ?
The derivative of f(x)=((x+9))/(x^2-7x+1) is (-x^2-18x+64)/((x^2-7x+1)^2)
What is the first derivative of f(x)=((x+9))/(x^2-7x+1) ?
The first derivative of f(x)=((x+9))/(x^2-7x+1) is (-x^2-18x+64)/((x^2-7x+1)^2)