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

The derivative of (4x^3+1/(x^2-x+3)) is (4x^4-8x^3+36x^2-2x+1)/((x^2-x+3)^2)

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

The first derivative of (4x^3+1/(x^2-x+3)) is (4x^4-8x^3+36x^2-2x+1)/((x^2-x+3)^2)

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

\frac{4 x ^{4} - 8 x ^{3} + 36 x ^{2} - 2 x + 1}{( x ^{2} - x + 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{3} + 1) = 12 x^{2}

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

= \frac{12 x ^{2} ( x ^{2} - x + 3 ) - ( 2 x - 1 ) ( 4 x ^{3} + 1 )}{( x ^{2} - x + 3 ) ^{2}}

Expand

12 x^{2} (x^{2} - x + 3) - (2 x - 1) (4 x^{3} + 1) : 4 x^{4} - 8 x^{3} + 36 x^{2} - 2 x + 1

= \frac{4 x ^{4} - 8 x ^{3} + 36 x ^{2} - 2 x + 1}{( x ^{2} - x + 3 ) ^{2}}

\frac{d y}{d x} - \frac{y}{x} = \frac{x ^{2} + y ^{2}}{x}

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

\frac{d}{d x} (x + 7)

\frac{d y}{d x}, x^{2} + 5 x + 9 x y - y^{2} = 4

1 + 6 x^{\frac{3}{2}}

What is the derivative of (4x^3+1/(x^2-x+3)) ?
The derivative of (4x^3+1/(x^2-x+3)) is (4x^4-8x^3+36x^2-2x+1)/((x^2-x+3)^2)
What is the first derivative of (4x^3+1/(x^2-x+3)) ?
The first derivative of (4x^3+1/(x^2-x+3)) is (4x^4-8x^3+36x^2-2x+1)/((x^2-x+3)^2)