What is (\partial)/(\partial x)((20)/(8+x^2+y^2)) ?

The answer to (\partial)/(\partial x)((20)/(8+x^2+y^2)) is -(40x)/((8+x^2+y^2)^2)

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(\partial)/(\partial x)((20)/(8+x^2+y^2))

\frac{\partial}{\partial x} (\frac{20}{8 + x ^{2} + y ^{2}})

- \frac{40 x}{( 8 + x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{20}{8 + x ^{2} + y ^{2}})

Treat

y

as a constant

Take the constant out:

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

= 20 \frac{\partial}{\partial x} (\frac{1}{8 + x ^{2} + y ^{2}})

Apply exponent rule:

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

= 20 \frac{\partial}{\partial x} ((8 + x^{2} + y^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( 8 + x ^{2} + y ^{2} ) ^{2}} \frac{\partial}{\partial x} (8 + x^{2} + y^{2})

= - \frac{1}{( 8 + x ^{2} + y ^{2} ) ^{2}} \frac{\partial}{\partial x} (8 + x^{2} + y^{2})

\frac{\partial}{\partial x} (8 + x^{2} + y^{2}) = 2 x

= 20 (- \frac{1}{( 8 + x ^{2} + y ^{2} ) ^{2}} \cdot 2 x)

Simplify

20 (- \frac{1}{( 8 + x ^{2} + y ^{2} ) ^{2}} \cdot 2 x) : - \frac{40 x}{( 8 + x ^{2} + y ^{2} ) ^{2}}

= - \frac{40 x}{( 8 + x ^{2} + y ^{2} ) ^{2}}

\frac{\partial}{\partial t} (\frac{8.31 t}{V ( t )})

x \to 2 lim (\frac{2 x - 6}{x - 1})

t e^{8 t}

x \to 9 - lim (\frac{5 x}{x - 9})

\frac{d}{d x} (- 4 x^{4} + 8 x^{3})

What is (\partial)/(\partial x)((20)/(8+x^2+y^2)) ?
The answer to (\partial)/(\partial x)((20)/(8+x^2+y^2)) is -(40x)/((8+x^2+y^2)^2)