What is the derivative of (4x^2-3/(x-1)) ?

The derivative of (4x^2-3/(x-1)) is (4x^2-8x+3)/((x-1)^2)

What is the first derivative of (4x^2-3/(x-1)) ?

The first derivative of (4x^2-3/(x-1)) is (4x^2-8x+3)/((x-1)^2)

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

\frac{4 x ^{2} - 8 x + 3}{( x - 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{2} - 3) = 8 x

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

= \frac{8 x ( x - 1 ) - 1 \cdot ( 4 x ^{2} - 3 )}{( x - 1 ) ^{2}}

Simplify

\frac{8 x ( x - 1 ) - 1 \cdot ( 4 x ^{2} - 3 )}{( x - 1 ) ^{2}} : \frac{4 x ^{2} - 8 x + 3}{( x - 1 ) ^{2}}

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

\frac{d}{d t} ((sin (t) + cos (t))^{2})

y = 5 x^{2} + \frac{6}{x}

y = \frac{8 t}{8 + t}

x \to 2 lim (2 x e^{x^{2}})

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

What is the derivative of (4x^2-3/(x-1)) ?
The derivative of (4x^2-3/(x-1)) is (4x^2-8x+3)/((x-1)^2)
What is the first derivative of (4x^2-3/(x-1)) ?
The first derivative of (4x^2-3/(x-1)) is (4x^2-8x+3)/((x-1)^2)