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

The derivative of f(x)=(x^3+2x^2+x-1)/x is (2x^3+2x^2+1)/(x^2)

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

The first derivative of f(x)=(x^3+2x^2+x-1)/x is (2x^3+2x^2+1)/(x^2)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d x}{d x} = 1

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

Simplify

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

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

n = 0 \sum \infty (- 1)^{n} (2)^{n}

f (x) = x^{2} - 2, - 3, 1

x y^{^{'''}} + 2 y^{^{''}} = 6 x

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

\frac{d}{d x} (\frac{x ^{2} + 5}{x ^{2} - 1})

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