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

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

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

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

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

\frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{8 x ^{3} - 3 x ^{2} + 6 x - 3}{2 x ^{4} - x ^{3} + 3 x ^{2} - 3 x}

f (x) = ln (1)

\int x e^{(3 x)} d x

\int \frac{15}{4 + 5 x} d x

\int_{- 3}^{4} y - (y^{2} - 12) d y

(- 1, 6), (7, - 2)

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