What is the derivative of y=((3x^2+4)(2x+1))/(5x-9) ?

The derivative of y=((3x^2+4)(2x+1))/(5x-9) is (60x^3-147x^2-54x-92)/((5x-9)^2)

What is the first derivative of y=((3x^2+4)(2x+1))/(5x-9) ?

The first derivative of y=((3x^2+4)(2x+1))/(5x-9) is (60x^3-147x^2-54x-92)/((5x-9)^2)

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derivative of

y = \frac{( 3 x ^{2} + 4 ) ( 2 x + 1 )}{5 x - 9}

\frac{60 x ^{3} - 147 x ^{2} - 54 x - 92}{( 5 x - 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( 3 x ^{2} + 4 ) ( 2 x + 1 )}{5 x - 9})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( 3 x ^{2} + 4 ) ( 2 x + 1 ) ) ( 5 x - 9 ) - \frac{d}{d x} ( 5 x - 9 ) ( 3 x ^{2} + 4 ) ( 2 x + 1 )}{( 5 x - 9 ) ^{2}}

\frac{d}{d x} ((3 x^{2} + 4) (2 x + 1)) = 18 x^{2} + 6 x + 8

\frac{d}{d x} (5 x - 9) = 5

= \frac{( 18 x ^{2} + 6 x + 8 ) ( 5 x - 9 ) - 5 ( 3 x ^{2} + 4 ) ( 2 x + 1 )}{( 5 x - 9 ) ^{2}}

Expand

(18 x^{2} + 6 x + 8) (5 x - 9) - 5 (3 x^{2} + 4) (2 x + 1) : 60 x^{3} - 147 x^{2} - 54 x - 92

= \frac{60 x ^{3} - 147 x ^{2} - 54 x - 92}{( 5 x - 9 ) ^{2}}

f (z) = \frac{z - 8}{z + 8}

x \to 3 lim (\frac{5}{x ^{2} - 9})

x \to 0 + lim (x^{4 x})

\frac{d y}{d x} = \frac{1}{5} x

f (x) = x^{2} - 4, at x = 3

What is the derivative of y=((3x^2+4)(2x+1))/(5x-9) ?
The derivative of y=((3x^2+4)(2x+1))/(5x-9) is (60x^3-147x^2-54x-92)/((5x-9)^2)
What is the first derivative of y=((3x^2+4)(2x+1))/(5x-9) ?
The first derivative of y=((3x^2+4)(2x+1))/(5x-9) is (60x^3-147x^2-54x-92)/((5x-9)^2)