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

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

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

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

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derivative of

f (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}

y^{^{'}} tan^{- 1} (x) - y (1 + x^{2})^{- 1} = 0

\frac{d}{d x} (cos (x) + 2 sin (x) cos (x))

\int \frac{w}{1 + w ^{2}} d w

\frac{x}{x ^{2} + 14 x + 45}

\frac{d}{d x} (e^{(- 5 x^{3} - 2 x^{2})})

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