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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Simplify

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

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

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n = 1 \sum \infty \frac{n !}{10 0 ^{n}}

\frac{d}{d x} (sin (5 x) + tan (3 x))

x^{^{''}} + 2 x^{^{'}} + 10 = e^{- t}, x (0) = 0, x^{^{'}} (0) = 0

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