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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{3} x^{2} 9 - x^{2} d x

\int \frac{1}{t ^{2} 256 - t ^{2}} d t

n = 0 \sum \infty (\frac{n x}{6})^{n}

x \to + 2 lim (2 - x)

f (x) = (2 t - 1) (3 t - 4)^{- 1}

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