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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d x}{d y} = - \frac{8 y ^{2} + 10 x y}{5 y ^{2} + 2 x}

\frac{\partial}{\partial x} (\frac{x ^{2}}{2} + y^{4})

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

\frac{\partial}{\partial x} (1 x)

(1 + y) \frac{d y}{d t} = 6 t, y (2) = 5

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