What is (\partial)/(\partial y)(sqrt((x+2)/(y+2))) ?

The answer to (\partial)/(\partial y)(sqrt((x+2)/(y+2))) is -(sqrt(x+2))/(2(y+2)^{3/2)}

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(\partial)/(\partial y)(sqrt((x+2)/(y+2)))

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x^{3} (x - 8)^{2})

\frac{d}{d x} (\frac{1}{x - y})

\frac{d y}{d x} + \frac{1}{x} y = 0

\int 2 cos (\frac{x}{3}) d x

f (x) = x^{2} - 6 x - 7

What is (\partial)/(\partial y)(sqrt((x+2)/(y+2))) ?
The answer to (\partial)/(\partial y)(sqrt((x+2)/(y+2))) is -(sqrt(x+2))/(2(y+2)^{3/2)}