What is the derivative of (3x^2-9x+1/(7x+8)) ?

The derivative of (3x^2-9x+1/(7x+8)) is (21x^2+48x-79)/((7x+8)^2)

What is the first derivative of (3x^2-9x+1/(7x+8)) ?

The first derivative of (3x^2-9x+1/(7x+8)) is (21x^2+48x-79)/((7x+8)^2)

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\frac{d}{d x} (\frac{3 x ^{2} - 9 x + 1}{7 x + 8})

\frac{21 x ^{2} + 48 x - 79}{( 7 x + 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{2} - 9 x + 1}{7 x + 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x ^{2} - 9 x + 1 ) ( 7 x + 8 ) - \frac{d}{d x} ( 7 x + 8 ) ( 3 x ^{2} - 9 x + 1 )}{( 7 x + 8 ) ^{2}}

\frac{d}{d x} (3 x^{2} - 9 x + 1) = 6 x - 9

\frac{d}{d x} (7 x + 8) = 7

= \frac{( 6 x - 9 ) ( 7 x + 8 ) - 7 ( 3 x ^{2} - 9 x + 1 )}{( 7 x + 8 ) ^{2}}

Expand

(6 x - 9) (7 x + 8) - 7 (3 x^{2} - 9 x + 1) : 21 x^{2} + 48 x - 79

= \frac{21 x ^{2} + 48 x - 79}{( 7 x + 8 ) ^{2}}

\int \frac{18 - 18 x}{1 - x} d x

x \to 0 + lim ((5 x)^{2 x})

f (x) = x^{2} + x + 4

\frac{\partial}{\partial u} (2 u + v)

\int 5 x^{3} 3 (2 x^{4} - 4)^{2} d x

What is the derivative of (3x^2-9x+1/(7x+8)) ?
The derivative of (3x^2-9x+1/(7x+8)) is (21x^2+48x-79)/((7x+8)^2)
What is the first derivative of (3x^2-9x+1/(7x+8)) ?
The first derivative of (3x^2-9x+1/(7x+8)) is (21x^2+48x-79)/((7x+8)^2)