What is the derivative of (sqrt(x)/(x^2+1)) ?

The derivative of (sqrt(x)/(x^2+1)) is (-3x^2+1)/(2sqrt(x)(x^2+1)^2)

What is the first derivative of (sqrt(x)/(x^2+1)) ?

The first derivative of (sqrt(x)/(x^2+1)) is (-3x^2+1)/(2sqrt(x)(x^2+1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f^{^{'}} (x) = - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e}

f (x) = x^{2} - 6 x + 8, g (x) = - 2 x + 8

y^{^{''}} + 8 y^{^{'}} + 16 = 0

w^{\frac{3}{2}} (w + c e^{w})

\frac{d}{d x} (ln (x^{3} + 3))

What is the derivative of (sqrt(x)/(x^2+1)) ?
The derivative of (sqrt(x)/(x^2+1)) is (-3x^2+1)/(2sqrt(x)(x^2+1)^2)
What is the first derivative of (sqrt(x)/(x^2+1)) ?
The first derivative of (sqrt(x)/(x^2+1)) is (-3x^2+1)/(2sqrt(x)(x^2+1)^2)