What is the derivative of ((x^2+4x+3)/(sqrt(x))) ?

The derivative of ((x^2+4x+3)/(sqrt(x))) is (3x^2+4x-3)/(2xsqrt(x))

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

The first derivative of ((x^2+4x+3)/(sqrt(x))) is (3x^2+4x-3)/(2xsqrt(x))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x) = \frac{1}{2 x}

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

Simplify

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

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

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

n = 1 \sum \infty 8 (- 1)^{n} n x^{n}

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

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

\int 10 x ln (3 x) d x

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