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

The derivative of f(x)=(x^2-3x+4)/(x+2) is (x^2+4x-10)/((x+2)^2)

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

The first derivative of f(x)=(x^2-3x+4)/(x+2) is (x^2+4x-10)/((x+2)^2)

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derivative of

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

\frac{x ^{2} + 4 x - 10}{( x + 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 x - 3 ) ( x + 2 ) - 1 \cdot ( x ^{2} - 3 x + 4 )}{( x + 2 ) ^{2}} : \frac{x ^{2} + 4 x - 10}{( x + 2 ) ^{2}}

= \frac{x ^{2} + 4 x - 10}{( x + 2 ) ^{2}}

1 + x - x^{2} - x^{3}, 1 - x

\frac{d}{d x} (\frac{6}{2 x + 1})

\frac{d}{d x} (\frac{x + 1}{x + 6})^{8}

y^{^{'}} = \frac{1}{( t - 1 ) ( y + 1 )}

f (x) = x^{3} (x - 6)^{2}

What is the derivative of f(x)=(x^2-3x+4)/(x+2) ?
The derivative of f(x)=(x^2-3x+4)/(x+2) is (x^2+4x-10)/((x+2)^2)
What is the first derivative of f(x)=(x^2-3x+4)/(x+2) ?
The first derivative of f(x)=(x^2-3x+4)/(x+2) is (x^2+4x-10)/((x+2)^2)