What is (\partial)/(\partial x)(2/(xz-z^4+x^2)) ?

The answer to (\partial)/(\partial x)(2/(xz-z^4+x^2)) is -(2(2x+z))/((xz-z^4+x^2)^2)

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(\partial)/(\partial x)(2/(xz-z^4+x^2))

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

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

Solution steps

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

Treat

z

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

2 (- \frac{1}{( x z - z ^{4} + x ^{2} ) ^{2}} (2 x + z)) : - \frac{2 ( 2 x + z )}{( x z - z ^{4} + x ^{2} ) ^{2}}

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

t^{2} \frac{d y}{d t} + y = 0, y (6) = - 3, t > 0

x \to 3 + lim (3 ln (x - 2))

\int cos (3 x^{2} + 5) x d x

f (x) = x^{4}, (- 2, 16)

\int 7 x (e^{6 x^{2}} - 1) d x

What is (\partial)/(\partial x)(2/(xz-z^4+x^2)) ?
The answer to (\partial)/(\partial x)(2/(xz-z^4+x^2)) is -(2(2x+z))/((xz-z^4+x^2)^2)