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derivative of y=(x^4-3x^3-6x^2-3x+2)/x

Solution

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

Solution

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

Solution steps

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

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2}{x} d x

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

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

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Frequently Asked Questions (FAQ)

What is the solution for derivative of y=(x^4-3x^3-6x^2-3x+2)/x ?
The solution for derivative of y=(x^4-3x^3-6x^2-3x+2)/x is y=(x^4)/4-x^3-3x^2-3x+2ln(x)+c_{1}