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

The derivative of (4x^2+2x+4)/(sqrt(x)) is (6x^2+x-2)/(xsqrt(x))

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

The first derivative of (4x^2+2x+4)/(sqrt(x)) is (6x^2+x-2)/(xsqrt(x))

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derivative of

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

\frac{6 x ^{2} + x - 2}{x x}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{2} + 2 x + 4) = 8 x + 2

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

= \frac{( 8 x + 2 ) x - \frac{1}{2 x} ( 4 x ^{2} + 2 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 8 x + 2 ) x - \frac{1}{2 x} ( 4 x ^{2} + 2 x + 4 )}{( x ) ^{2}} : \frac{6 x ^{2} + x - 2}{x x}

= \frac{6 x ^{2} + x - 2}{x x}

\frac{d y}{d x} = e^{4 x} + 5 y

x \to 0 lim (5 x^{4})

\int_{0}^{1} 7 x x^{2} + 16 d x

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

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

What is the derivative of (4x^2+2x+4)/(sqrt(x)) ?
The derivative of (4x^2+2x+4)/(sqrt(x)) is (6x^2+x-2)/(xsqrt(x))
What is the first derivative of (4x^2+2x+4)/(sqrt(x)) ?
The first derivative of (4x^2+2x+4)/(sqrt(x)) is (6x^2+x-2)/(xsqrt(x))