What is (\partial)/(\partial x)(1/((1/x+1/y))) ?

The answer to (\partial)/(\partial x)(1/((1/x+1/y))) is (y^2)/((y+x)^2)

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(\partial)/(\partial x)(1/((1/x+1/y)))

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{6 x ^{3}}{16 + x ^{2}} d x

y^{^{'}} = 2 y + x^{2} + 9

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

f (x) = x^{2} + a x + 2, at x = 1

\int - 6 ln (x^{2} - 1) d x

What is (\partial)/(\partial x)(1/((1/x+1/y))) ?
The answer to (\partial)/(\partial x)(1/((1/x+1/y))) is (y^2)/((y+x)^2)