What is (\partial)/(\partial x)((100)/(4x^2+y^2)) ?

The answer to (\partial)/(\partial x)((100)/(4x^2+y^2)) is -(800x)/((4x^2+y^2)^2)

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(\partial)/(\partial x)((100)/(4x^2+y^2))

\frac{\partial}{\partial x} (\frac{100}{4 x ^{2} + y ^{2}})

- \frac{800 x}{( 4 x ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{100}{4 x ^{2} + y ^{2}})

Treat

y

as a constant

Take the constant out:

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

= 100 \frac{\partial}{\partial x} (\frac{1}{4 x ^{2} + y ^{2}})

Apply exponent rule:

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

= 100 \frac{\partial}{\partial x} ((4 x^{2} + y^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( 4 x ^{2} + y ^{2} ) ^{2}} \frac{\partial}{\partial x} (4 x^{2} + y^{2})

= - \frac{1}{( 4 x ^{2} + y ^{2} ) ^{2}} \frac{\partial}{\partial x} (4 x^{2} + y^{2})

\frac{\partial}{\partial x} (4 x^{2} + y^{2}) = 8 x

= 100 (- \frac{1}{( 4 x ^{2} + y ^{2} ) ^{2}} \cdot 8 x)

Simplify

100 (- \frac{1}{( 4 x ^{2} + y ^{2} ) ^{2}} \cdot 8 x) : - \frac{800 x}{( 4 x ^{2} + y ^{2} ) ^{2}}

= - \frac{800 x}{( 4 x ^{2} + y ^{2} ) ^{2}}

\frac{d}{d x} (x - 2 x^{2})

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What is (\partial)/(\partial x)((100)/(4x^2+y^2)) ?
The answer to (\partial)/(\partial x)((100)/(4x^2+y^2)) is -(800x)/((4x^2+y^2)^2)