What is the derivative of (4sqrt(x)/(3x^2+6x-8)) ?

The derivative of (4sqrt(x)/(3x^2+6x-8)) is (2(-9x^2-6x-8))/(sqrt(x)(3x^2+6x-8)^2)

What is the first derivative of (4sqrt(x)/(3x^2+6x-8)) ?

The first derivative of (4sqrt(x)/(3x^2+6x-8)) is (2(-9x^2-6x-8))/(sqrt(x)(3x^2+6x-8)^2)

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\frac{d}{d x} (\frac{4 x}{3 x ^{2} + 6 x - 8})

\frac{2 ( - 9 x ^{2} - 6 x - 8 )}{x ( 3 x ^{2} + 6 x - 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4 x}{3 x ^{2} + 6 x - 8})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (\frac{x}{3 x ^{2} + 6 x - 8})

Apply the Quotient Rule:

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

= 4 \cdot \frac{\frac{d}{d x} ( x ) ( 3 x ^{2} + 6 x - 8 ) - \frac{d}{d x} ( 3 x ^{2} + 6 x - 8 ) x}{( 3 x ^{2} + 6 x - 8 ) ^{2}}

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

\frac{d}{d x} (3 x^{2} + 6 x - 8) = 6 x + 6

= 4 \cdot \frac{\frac{1}{2 x} ( 3 x ^{2} + 6 x - 8 ) - ( 6 x + 6 ) x}{( 3 x ^{2} + 6 x - 8 ) ^{2}}

Simplify

4 \cdot \frac{\frac{1}{2 x} ( 3 x ^{2} + 6 x - 8 ) - ( 6 x + 6 ) x}{( 3 x ^{2} + 6 x - 8 ) ^{2}} : \frac{2 ( - 9 x ^{2} - 6 x - 8 )}{x ( 3 x ^{2} + 6 x - 8 ) ^{2}}

= \frac{2 ( - 9 x ^{2} - 6 x - 8 )}{x ( 3 x ^{2} + 6 x - 8 ) ^{2}}

\int 17 d x

\frac{\partial}{\partial y} ((x - 1) y^{2} e^{x y})

y^{^{'}} = 5 y + e^{- 2 x} y^{- 2}

\frac{\partial}{\partial x} (cos (5 x - 4 y^{2}))

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

What is the derivative of (4sqrt(x)/(3x^2+6x-8)) ?
The derivative of (4sqrt(x)/(3x^2+6x-8)) is (2(-9x^2-6x-8))/(sqrt(x)(3x^2+6x-8)^2)
What is the first derivative of (4sqrt(x)/(3x^2+6x-8)) ?
The first derivative of (4sqrt(x)/(3x^2+6x-8)) is (2(-9x^2-6x-8))/(sqrt(x)(3x^2+6x-8)^2)