What is (\partial)/(\partial x)(2/(2x-2z+1)) ?

The answer to (\partial)/(\partial x)(2/(2x-2z+1)) is -4/((2x-2z+1)^2)

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(\partial)/(\partial x)(2/(2x-2z+1))

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

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

Solution steps

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

Treat

z

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \cdot 3 x^{2} + 5

x^{2}, x, - 1, 1

x^{2} - 2, x, - 2, 2

y^{^{'}} = \frac{4 x y}{2 + x ^{2}}

x \to - 4 lim (- x^{2} + 4 x - 9)

What is (\partial)/(\partial x)(2/(2x-2z+1)) ?
The answer to (\partial)/(\partial x)(2/(2x-2z+1)) is -4/((2x-2z+1)^2)