What is the derivative of (8x^2+4x+4)/(sqrt(x)) ?

The derivative of (8x^2+4x+4)/(sqrt(x)) is (12x^2+2x-2)/(xsqrt(x))

What is the first derivative of (8x^2+4x+4)/(sqrt(x)) ?

The first derivative of (8x^2+4x+4)/(sqrt(x)) is (12x^2+2x-2)/(xsqrt(x))

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derivative of

\frac{8 x ^{2} + 4 x + 4}{x}

\frac{12 x ^{2} + 2 x - 2}{x x}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (8 x^{2} + 4 x + 4) = 16 x + 4

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

= \frac{( 16 x + 4 ) x - \frac{1}{2 x} ( 8 x ^{2} + 4 x + 4 )}{( x ) ^{2}}

Simplify

\frac{( 16 x + 4 ) x - \frac{1}{2 x} ( 8 x ^{2} + 4 x + 4 )}{( x ) ^{2}} : \frac{12 x ^{2} + 2 x - 2}{x x}

= \frac{12 x ^{2} + 2 x - 2}{x x}

x = y^{2} + 6 y, x = y

\int_{0}^{x} 4 tan (t) d t

y^{^{''}} + 9 y = e^{3 x}

\frac{\partial}{\partial x} (sin^{2} (x y))

121 y^{^{''}} + 44 y^{^{'}} + 4 y = 0

What is the derivative of (8x^2+4x+4)/(sqrt(x)) ?
The derivative of (8x^2+4x+4)/(sqrt(x)) is (12x^2+2x-2)/(xsqrt(x))
What is the first derivative of (8x^2+4x+4)/(sqrt(x)) ?
The first derivative of (8x^2+4x+4)/(sqrt(x)) is (12x^2+2x-2)/(xsqrt(x))