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

The derivative of (8x^2+8+4)/(sqrt(x)) is (12x^2-6)/(xsqrt(x))

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

The first derivative of (8x^2+8+4)/(sqrt(x)) is (12x^2-6)/(xsqrt(x))

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derivative of

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

\frac{12 x ^{2} - 6}{x x}

Solution steps

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

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

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

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

Simplify

\frac{16 x x - \frac{1}{2 x} ( 8 x ^{2} + 8 + 4 )}{( x ) ^{2}} : \frac{12 x ^{2} - 6}{x x}

= \frac{12 x ^{2} - 6}{x x}

\int (x + 5) 10 x + x^{2} d x

f (x) = x + \frac{4}{x}, (4, 5)

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

\frac{d}{d x} (- \frac{36}{( 1 + 6 x ) ^{2}})

\frac{d y}{d x} = \frac{9 x ^{2} - 6}{x ^{2}}

What is the derivative of (8x^2+8+4)/(sqrt(x)) ?
The derivative of (8x^2+8+4)/(sqrt(x)) is (12x^2-6)/(xsqrt(x))
What is the first derivative of (8x^2+8+4)/(sqrt(x)) ?
The first derivative of (8x^2+8+4)/(sqrt(x)) is (12x^2-6)/(xsqrt(x))