What is (\partial)/(\partial x)(2/(2x+4y+1)) ?

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

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(\partial)/(\partial x)(2/(2x+4y+1))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (3 sin (x) cos^{2} (x) + sin^{3} (x))

(2 x y^{2} + 2 y) d x + (2 x^{2} y + 2 x) d y = 0

\frac{\partial}{\partial x} (100 e^{(- x - 2 y)})

x \to 1 lim (g (x) (f (x)))

x \to 4 lim (\frac{( 3 x ^{2} - 17 x + 20 )}{4 x ^{2} - 25 x + 36})

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