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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (y - 1) = 1

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

Simplify

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

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

\int_{0}^{1} \frac{1}{x ( 1 + x )} d x

\int y y + 1 d y

d er i v a t i v e 4 tan^{2} (3 x)

\int - ln (x) d x

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

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