What is the derivative of f(x)=(6x-8)/(x^2-6x+7) ?

The derivative of f(x)=(6x-8)/(x^2-6x+7) is (-6x^2+16x-6)/((x^2-6x+7)^2)

What is the first derivative of f(x)=(6x-8)/(x^2-6x+7) ?

The first derivative of f(x)=(6x-8)/(x^2-6x+7) is (-6x^2+16x-6)/((x^2-6x+7)^2)

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derivative of

f (x) = \frac{6 x - 8}{x ^{2} - 6 x + 7}

\frac{- 6 x ^{2} + 16 x - 6}{( x ^{2} - 6 x + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x - 8}{x ^{2} - 6 x + 7})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 x - 8 ) ( x ^{2} - 6 x + 7 ) - \frac{d}{d x} ( x ^{2} - 6 x + 7 ) ( 6 x - 8 )}{( x ^{2} - 6 x + 7 ) ^{2}}

\frac{d}{d x} (6 x - 8) = 6

\frac{d}{d x} (x^{2} - 6 x + 7) = 2 x - 6

= \frac{6 ( x ^{2} - 6 x + 7 ) - ( 2 x - 6 ) ( 6 x - 8 )}{( x ^{2} - 6 x + 7 ) ^{2}}

Expand

6 (x^{2} - 6 x + 7) - (2 x - 6) (6 x - 8) : - 6 x^{2} + 16 x - 6

= \frac{- 6 x ^{2} + 16 x - 6}{( x ^{2} - 6 x + 7 ) ^{2}}

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

\int 10 sin^{3} (x) cos^{2} (x) d x

y = (6 - x^{2}) (x^{3} - 3 x + 4)

\int ∣ u + 1 ∣ d u

\frac{\partial}{\partial t} (r sin (t) sin (x))

What is the derivative of f(x)=(6x-8)/(x^2-6x+7) ?
The derivative of f(x)=(6x-8)/(x^2-6x+7) is (-6x^2+16x-6)/((x^2-6x+7)^2)
What is the first derivative of f(x)=(6x-8)/(x^2-6x+7) ?
The first derivative of f(x)=(6x-8)/(x^2-6x+7) is (-6x^2+16x-6)/((x^2-6x+7)^2)