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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to - 1 lim (x^{4} - x^{3} + 2 x)

\frac{d}{d x} (e^{- 2 x} (a cos (x) + b sin (x)))

x \to 0 + lim (\frac{6}{sin ( x )})

\int \frac{1}{v ^{2} + 8 v - 9} d v

\int - \frac{2}{x + 1} d x

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