What is (\partial)/(\partial y)(-5/(1+x^2-y)) ?

The answer to (\partial)/(\partial y)(-5/(1+x^2-y)) is -5/((1+x^2-y)^2)

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(\partial)/(\partial y)(-5/(1+x^2-y))

\frac{\partial}{\partial y} (- \frac{5}{1 + x ^{2} - y})

- \frac{5}{( 1 + x ^{2} - y ) ^{2}}

Solution steps

\frac{\partial}{\partial y} (- \frac{5}{1 + x ^{2} - y})

Treat

x

as a constant

Take the constant out:

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

= - 5 \frac{\partial}{\partial y} (\frac{1}{1 + x ^{2} - y})

Apply exponent rule:

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

= - 5 \frac{\partial}{\partial y} ((1 + x^{2} - y)^{- 1})

Apply the chain rule:

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

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

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

= - 5 (- \frac{1}{( 1 + x ^{2} - y ) ^{2}} (- 1))

Simplify

- 5 (- \frac{1}{( 1 + x ^{2} - y ) ^{2}} (- 1)) : - \frac{5}{( 1 + x ^{2} - y ) ^{2}}

= - \frac{5}{( 1 + x ^{2} - y ) ^{2}}

f (x) = x^{3}, at x = - 2

\int 21 sec^{4} (x) d x

\frac{d}{d x} (- 3 e^{- x})

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

\frac{d}{d x} (\frac{0.6 e ^{x}}{cos ( x )})

What is (\partial)/(\partial y)(-5/(1+x^2-y)) ?
The answer to (\partial)/(\partial y)(-5/(1+x^2-y)) is -5/((1+x^2-y)^2)