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

The derivative of ((8x^2-2x-9))/((2x-3)) is (16x^2-48x+24)/((2x-3)^2)

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

The first derivative of ((8x^2-2x-9))/((2x-3)) is (16x^2-48x+24)/((2x-3)^2)

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derivative of

\frac{( 8 x ^{2} - 2 x - 9 )}{( 2 x - 3 )}

\frac{16 x ^{2} - 48 x + 24}{( 2 x - 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (8 x^{2} - 2 x - 9) = 16 x - 2

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

= \frac{( 16 x - 2 ) ( 2 x - 3 ) - 2 ( 8 x ^{2} - 2 x - 9 )}{( 2 x - 3 ) ^{2}}

Expand

(16 x - 2) (2 x - 3) - 2 (8 x^{2} - 2 x - 9) : 16 x^{2} - 48 x + 24

= \frac{16 x ^{2} - 48 x + 24}{( 2 x - 3 ) ^{2}}

y^{^{'}} (x^{2} + 2 x + 5) = (1 + y^{2}) (x + 1)

\int e^{5 x^{2}} d x

f (x) = ln (1 - x^{2} + 2 x^{4}), at x = 1

\int \frac{1}{( 3 x - 2 ) ^{2}} d x

n = 1 \sum \infty \frac{4}{n ( n + 3 )}

What is the derivative of ((8x^2-2x-9))/((2x-3)) ?
The derivative of ((8x^2-2x-9))/((2x-3)) is (16x^2-48x+24)/((2x-3)^2)
What is the first derivative of ((8x^2-2x-9))/((2x-3)) ?
The first derivative of ((8x^2-2x-9))/((2x-3)) is (16x^2-48x+24)/((2x-3)^2)