What is the derivative of f(x)=(x^2-6x+12)/(x-4) ?

The derivative of f(x)=(x^2-6x+12)/(x-4) is (x^2-8x+12)/((x-4)^2)

What is the first derivative of f(x)=(x^2-6x+12)/(x-4) ?

The first derivative of f(x)=(x^2-6x+12)/(x-4) is (x^2-8x+12)/((x-4)^2)

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derivative of

f (x) = \frac{x ^{2} - 6 x + 12}{x - 4}

\frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 6 x + 12}{x - 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 6 x + 12 ) ( x - 4 ) - \frac{d}{d x} ( x - 4 ) ( x ^{2} - 6 x + 12 )}{( x - 4 ) ^{2}}

\frac{d}{d x} (x^{2} - 6 x + 12) = 2 x - 6

\frac{d}{d x} (x - 4) = 1

= \frac{( 2 x - 6 ) ( x - 4 ) - 1 \cdot ( x ^{2} - 6 x + 12 )}{( x - 4 ) ^{2}}

Simplify

\frac{( 2 x - 6 ) ( x - 4 ) - 1 \cdot ( x ^{2} - 6 x + 12 )}{( x - 4 ) ^{2}} : \frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}}

= \frac{x ^{2} - 8 x + 12}{( x - 4 ) ^{2}}

x^{2} - 3 x

\int_{0}^{3} \frac{1}{x ^{4}} d x

\int 18 cot^{4} (x) d x

\int y^{2} - 1 d y

f (x) = x^{3} - 2 x + 2, (4, 58)

What is the derivative of f(x)=(x^2-6x+12)/(x-4) ?
The derivative of f(x)=(x^2-6x+12)/(x-4) is (x^2-8x+12)/((x-4)^2)
What is the first derivative of f(x)=(x^2-6x+12)/(x-4) ?
The first derivative of f(x)=(x^2-6x+12)/(x-4) is (x^2-8x+12)/((x-4)^2)