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

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

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

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

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

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

Solution steps

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

\frac{d}{d x} (x^{2} + x - 12) = 2 x + 1

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

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

Simplify

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

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

x \to 0 lim (\frac{2 x ^{2} + x}{x})

y = \frac{x ^{2}}{5 - 8 x}

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

x \to 0 lim (x^{3} \cdot sin (\frac{1}{x}))

x \to 0 lim ((1 - 2 x)^{\frac{5}{x}})

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