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

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

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

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

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

\frac{6 x ^{2} + x - 4}{x x}

Solution steps

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

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

Simplify

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

= \frac{6 x ^{2} + x - 4}{x x}

\frac{d}{d x} (2 x^{- 2})

y^{^{''}} - 8 y^{^{'}} + 20 y = 0

x \to π lim (e^{c o s (3 x)})

\int_{1}^{2} x 5^{(x^{2})} d x

x \to 0 lim (\frac{3 x ^{2} - 2 x}{x})

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