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

The derivative of (2x^3+4/(x^2-4x+1)) is (2x^4-16x^3+6x^2-8x+16)/((x^2-4x+1)^2)

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

The first derivative of (2x^3+4/(x^2-4x+1)) is (2x^4-16x^3+6x^2-8x+16)/((x^2-4x+1)^2)

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

\frac{2 x ^{4} - 16 x ^{3} + 6 x ^{2} - 8 x + 16}{( x ^{2} - 4 x + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{3} + 4) = 6 x^{2}

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

= \frac{6 x ^{2} ( x ^{2} - 4 x + 1 ) - ( 2 x - 4 ) ( 2 x ^{3} + 4 )}{( x ^{2} - 4 x + 1 ) ^{2}}

Expand

6 x^{2} (x^{2} - 4 x + 1) - (2 x - 4) (2 x^{3} + 4) : 2 x^{4} - 16 x^{3} + 6 x^{2} - 8 x + 16

= \frac{2 x ^{4} - 16 x ^{3} + 6 x ^{2} - 8 x + 16}{( x ^{2} - 4 x + 1 ) ^{2}}

x \to 0 lim (ln (tan (7 x)))

\int \frac{16 x ^{2}}{16 x ^{2} + 25} d x

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

f (x) = arctan (\frac{1 - x}{1 + x})

\frac{x}{7 + x}

What is the derivative of (2x^3+4/(x^2-4x+1)) ?
The derivative of (2x^3+4/(x^2-4x+1)) is (2x^4-16x^3+6x^2-8x+16)/((x^2-4x+1)^2)
What is the first derivative of (2x^3+4/(x^2-4x+1)) ?
The first derivative of (2x^3+4/(x^2-4x+1)) is (2x^4-16x^3+6x^2-8x+16)/((x^2-4x+1)^2)