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}}

t \to \infty lim (Q (t) + (100 - Q (t)) e^{- k t})

g (t) = 4 t^{- \frac{1}{8}} + 3 ln (t) + ln (5)

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

\int sec^{2} (θ) tan^{3} (θ) d θ

\frac{d}{d x} (x^{(\frac{1}{5} l n (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)