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

The derivative of (6(x-3^2)/(3(x^2-3x-4))) is (2(3x^2-26x+51))/((x^2-3x-4)^2)

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

The first derivative of (6(x-3^2)/(3(x^2-3x-4))) is (2(3x^2-26x+51))/((x^2-3x-4)^2)

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

\frac{2 ( 3 x ^{2} - 26 x + 51 )}{( x ^{2} - 3 x - 4 ) ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{6}{3} \cdot \frac{2 ( x - 3 ) ( x ^{2} - 3 x - 4 ) - ( 2 x - 3 ) ( x - 3 ) ^{2}}{( x ^{2} - 3 x - 4 ) ^{2}} : \frac{2 ( 3 x ^{2} - 26 x + 51 )}{( x ^{2} - 3 x - 4 ) ^{2}}

= \frac{2 ( 3 x ^{2} - 26 x + 51 )}{( x ^{2} - 3 x - 4 ) ^{2}}

\frac{9 cos ( x )}{1 + sin ( x )}

\frac{1}{s - 3} - \frac{16}{s ^{2} + 9}

x \to 6 lim (ln (x - 6))

y = (3 x^{2} + 4) (3 + x^{3})^{5}

17 x, x^{2}

What is the derivative of (6(x-3^2)/(3(x^2-3x-4))) ?
The derivative of (6(x-3^2)/(3(x^2-3x-4))) is (2(3x^2-26x+51))/((x^2-3x-4)^2)
What is the first derivative of (6(x-3^2)/(3(x^2-3x-4))) ?
The first derivative of (6(x-3^2)/(3(x^2-3x-4))) is (2(3x^2-26x+51))/((x^2-3x-4)^2)