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

The derivative of (4x+6/(sqrt(x^2+3x+4))) is 7/((x^2+3x+4)sqrt(x^2+3x+4))

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

The first derivative of (4x+6/(sqrt(x^2+3x+4))) is 7/((x^2+3x+4)sqrt(x^2+3x+4))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x + 6) = 4

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

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

Simplify

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

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

\frac{2}{( s ^{2} + 1 ) ^{2}}

y = \frac{x + \frac{1}{x}}{2 x + \frac{1}{x}}

\int \frac{1}{x 12 - x ^{2}} d x

\int sec (7 w) tan (7 w) d w

f (x) = arctan (\frac{x - 1}{x + 1})

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