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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derivative of

\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 x 1 0^{x} d x

(\frac{9 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

x^{2} - 2 x + 4, 2

\int_{0}^{1} \frac{1}{1 - x} d x

P (x) = x^{3} - 2 x^{2} + 2 x - 1

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))