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

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

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

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

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derivative of

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

x + \frac{3}{4}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= x + \frac{3}{4}

x \to - 4 - lim (\frac{x + 2}{x + 4})

\int_{- 3}^{3} 2 - ∣ x ∣ d x

\int \frac{26}{( x - 1 ) ( x ^{2} + 25 )} d x

\int 4 x^{3} - \frac{1}{1 - x} d x

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

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