What is the d/(dy)(1/(sqrt(x^2+y^2))) ?

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

What is the first d/(dy)(1/(sqrt(x^2+y^2))) ?

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (1 - 4 x^{2})

\int 64 - t^{2} d t

\frac{d}{d x} (- 6 e^{- x})

\int \frac{8}{x ^{2} - 36} d x

\frac{d}{d x} (e^{6} + ln (9))

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