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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{d}{d x} (4^{l n (x)})

\int_{0}^{1} 2 π x (15 - 15 x) d x

\int \frac{x ^{4} - 3 x}{6 x} d x

\int \frac{1}{x} ln (x) d x

\int - (t - 6)^{2} d t

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