What is the derivative of (7x^2-6x+4/(3x+2)) ?

The derivative of (7x^2-6x+4/(3x+2)) is (21x^2+28x-24)/((3x+2)^2)

What is the first derivative of (7x^2-6x+4/(3x+2)) ?

The first derivative of (7x^2-6x+4/(3x+2)) is (21x^2+28x-24)/((3x+2)^2)

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\frac{d}{d x} (\frac{7 x ^{2} - 6 x + 4}{3 x + 2})

\frac{21 x ^{2} + 28 x - 24}{( 3 x + 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7 x ^{2} - 6 x + 4}{3 x + 2})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 7 x ^{2} - 6 x + 4 ) ( 3 x + 2 ) - \frac{d}{d x} ( 3 x + 2 ) ( 7 x ^{2} - 6 x + 4 )}{( 3 x + 2 ) ^{2}}

\frac{d}{d x} (7 x^{2} - 6 x + 4) = 14 x - 6

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

= \frac{( 14 x - 6 ) ( 3 x + 2 ) - 3 ( 7 x ^{2} - 6 x + 4 )}{( 3 x + 2 ) ^{2}}

Expand

(14 x - 6) (3 x + 2) - 3 (7 x^{2} - 6 x + 4) : 21 x^{2} + 28 x - 24

= \frac{21 x ^{2} + 28 x - 24}{( 3 x + 2 ) ^{2}}

\int (6 x + 2 y) d y

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

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

f (x) = 5 - x

x \to 3 - lim (ln (\frac{1}{3 - x}))

What is the derivative of (7x^2-6x+4/(3x+2)) ?
The derivative of (7x^2-6x+4/(3x+2)) is (21x^2+28x-24)/((3x+2)^2)
What is the first derivative of (7x^2-6x+4/(3x+2)) ?
The first derivative of (7x^2-6x+4/(3x+2)) is (21x^2+28x-24)/((3x+2)^2)