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

The derivative of sqrt(x/(3x^2+4)) is (-3x^2+4)/(2sqrt(x)(3x^2+4)^{3/2)}

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

The first derivative of sqrt(x/(3x^2+4)) is (-3x^2+4)/(2sqrt(x)(3x^2+4)^{3/2)}

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

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

Solution steps

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

Apply the chain rule:

\frac{1}{2 \frac{x}{3 x ^{2} + 4}} \frac{d}{d x} (\frac{x}{3 x ^{2} + 4})

= \frac{1}{2 \frac{x}{3 x ^{2} + 4}} \frac{d}{d x} (\frac{x}{3 x ^{2} + 4})

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

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

Simplify

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

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

f (x) = - e^{x} x + e^{x}

2 x y^{^{'}} + y = 6 x, y (4) = 22

\int \frac{1}{( x + 2 ) ( x + 4 ) ( x + 6 )} d x

\frac{d}{d x} (lo g_{e} (x^{x}))

cos^{2} (θ)

What is the derivative of sqrt(x/(3x^2+4)) ?
The derivative of sqrt(x/(3x^2+4)) is (-3x^2+4)/(2sqrt(x)(3x^2+4)^{3/2)}
What is the first derivative of sqrt(x/(3x^2+4)) ?
The first derivative of sqrt(x/(3x^2+4)) is (-3x^2+4)/(2sqrt(x)(3x^2+4)^{3/2)}