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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

d er i v a t i v e \frac{x + 1}{2 x - 1}

d er i v a t i v e sin (y^{2})

d er i v a t i v e y = sin^{2} (x) 0.2 x

d er i v a t i v e y = e^{3} x

d er i v a t i v e e^{2 x - 1}

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