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

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

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

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

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derivative of

y = \frac{1}{x ^{2} + 1}

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int - \frac{2}{9 - 9 x ^{2}} d x

x y^{^{'}} + 3 y = \frac{( 2 e ^{4 x} )}{x ^{2}}

d er i v a t i v e y = \frac{1}{( 4 x ^{2} + 3 ) ^{5}}

\int (2 x^{3} + 5 x^{2} - 2 x + 7) d x

\frac{\partial}{\partial y} (x y + ln (\frac{x}{y}))

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