What is the derivative of (6x^2+8x+8/(sqrt(x))) ?

The derivative of (6x^2+8x+8/(sqrt(x))) is (9x^2+4x-4)/(xsqrt(x))

What is the first derivative of (6x^2+8x+8/(sqrt(x))) ?

The first derivative of (6x^2+8x+8/(sqrt(x))) is (9x^2+4x-4)/(xsqrt(x))

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\frac{d}{d x} (\frac{6 x ^{2} + 8 x + 8}{x})

\frac{9 x ^{2} + 4 x - 4}{x x}

Solution steps

\frac{d}{d x} (\frac{6 x ^{2} + 8 x + 8}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 x ^{2} + 8 x + 8 ) x - \frac{d}{d x} ( x ) ( 6 x ^{2} + 8 x + 8 )}{( x ) ^{2}}

\frac{d}{d x} (6 x^{2} + 8 x + 8) = 12 x + 8

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

= \frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 8 )}{( x ) ^{2}}

Simplify

\frac{( 12 x + 8 ) x - \frac{1}{2 x} ( 6 x ^{2} + 8 x + 8 )}{( x ) ^{2}} : \frac{9 x ^{2} + 4 x - 4}{x x}

= \frac{9 x ^{2} + 4 x - 4}{x x}

\frac{d}{d x} (84 x^{6} + 12 x^{3} - 24 x^{2})

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

\int \frac{1}{( 4 + 9 x ^{2} )} d x

x^{2} y^{^{''}} + 3 x y^{^{'}} - 4 y = 0

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

What is the derivative of (6x^2+8x+8/(sqrt(x))) ?
The derivative of (6x^2+8x+8/(sqrt(x))) is (9x^2+4x-4)/(xsqrt(x))
What is the first derivative of (6x^2+8x+8/(sqrt(x))) ?
The first derivative of (6x^2+8x+8/(sqrt(x))) is (9x^2+4x-4)/(xsqrt(x))