What is the derivative of f(x)=(x^2-1)/(2x-3) ?

The derivative of f(x)=(x^2-1)/(2x-3) is (2x^2-6x+2)/((2x-3)^2)

What is the first derivative of f(x)=(x^2-1)/(2x-3) ?

The first derivative of f(x)=(x^2-1)/(2x-3) is (2x^2-6x+2)/((2x-3)^2)

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derivative of

f (x) = \frac{x ^{2} - 1}{2 x - 3}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

2 x (2 x - 3) - 2 (x^{2} - 1) : 2 x^{2} - 6 x + 2

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

\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} - 5 y = 0

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

\frac{\partial}{\partial x} (A)

x \to 1 lim (\frac{ln ( x )}{x})

- \frac{2}{x}

What is the derivative of f(x)=(x^2-1)/(2x-3) ?
The derivative of f(x)=(x^2-1)/(2x-3) is (2x^2-6x+2)/((2x-3)^2)
What is the first derivative of f(x)=(x^2-1)/(2x-3) ?
The first derivative of f(x)=(x^2-1)/(2x-3) is (2x^2-6x+2)/((2x-3)^2)