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

The derivative of-(4x-3/((2x^2-3x+1)^2)) is -(2(-12x^2+18x-7))/((2x^2-3x+1)^3)

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

The first derivative of-(4x-3/((2x^2-3x+1)^2)) is -(2(-12x^2+18x-7))/((2x^2-3x+1)^3)

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

- \frac{2 ( - 12 x ^{2} + 18 x - 7 )}{( 2 x ^{2} - 3 x + 1 ) ^{3}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

- \frac{4 ( 2 x ^{2} - 3 x + 1 ) ^{2} - 2 ( 2 x ^{2} - 3 x + 1 ) ( 4 x - 3 ) ( 4 x - 3 )}{( ( 2 x ^{2} - 3 x + 1 ) ^{2} ) ^{2}} : - \frac{2 ( - 12 x ^{2} + 18 x - 7 )}{( 2 x ^{2} - 3 x + 1 ) ^{3}}

= - \frac{2 ( - 12 x ^{2} + 18 x - 7 )}{( 2 x ^{2} - 3 x + 1 ) ^{3}}

\int_{- 5}^{5} 4 + ∣ x ∣ d x

x \to 3 + lim (x - 1)

\int sin (2 x) cos^{3} (2 x) d x

x \to 4 lim (\frac{2 x}{x - 4})

\frac{\partial}{\partial x} (sin (x y e^{x^{2}}))

What is the derivative of-(4x-3/((2x^2-3x+1)^2)) ?
The derivative of-(4x-3/((2x^2-3x+1)^2)) is -(2(-12x^2+18x-7))/((2x^2-3x+1)^3)
What is the first derivative of-(4x-3/((2x^2-3x+1)^2)) ?
The first derivative of-(4x-3/((2x^2-3x+1)^2)) is -(2(-12x^2+18x-7))/((2x^2-3x+1)^3)