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

The answer to (\partial)/(\partial y)(-(2x)/((y+2z)^2)) is (4x)/((y+2z)^3)

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(\partial)/(\partial y)(-(2x)/((y+2z)^2))

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

\frac{4 x}{( y + 2 z ) ^{3}}

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{2}{( y + 2 z ) ^{3}} \frac{\partial}{\partial y} (y + 2 z)

= - \frac{2}{( y + 2 z ) ^{3}} \frac{\partial}{\partial y} (y + 2 z)

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

= - 2 x (- \frac{2}{( y + 2 z ) ^{3}} \cdot 1)

Simplify

- 2 x (- \frac{2}{( y + 2 z ) ^{3}} \cdot 1) : \frac{4 x}{( y + 2 z ) ^{3}}

= \frac{4 x}{( y + 2 z ) ^{3}}

f (x) = x^{4} - 5 x^{2} + 4, at x = - 1

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

x \to π lim (3 sin (x))

\int cos (3 x) cos (4 x) d x

\frac{1}{( s + 3 ) ^{2} + 25}

What is (\partial)/(\partial y)(-(2x)/((y+2z)^2)) ?
The answer to (\partial)/(\partial y)(-(2x)/((y+2z)^2)) is (4x)/((y+2z)^3)