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

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

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

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

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derivative of

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = x^{5} + x^{3} - x + 10

5 x^{2} + 3 x + 5

4 x^{2} - 5 x - 1

\frac{d}{d y} (ln (y + 2) \cdot e^{2} x y)

\frac{( e ^{2} x )}{x}

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