What is (\partial)/(\partial x)((-8x+5)/(2y^2+1)) ?

The answer to (\partial)/(\partial x)((-8x+5)/(2y^2+1)) is -8/(2y^2+1)

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(\partial)/(\partial x)((-8x+5)/(2y^2+1))

\frac{\partial}{\partial x} (\frac{- 8 x + 5}{2 y ^{2} + 1})

- \frac{8}{2 y ^{2} + 1}

Solution steps

\frac{\partial}{\partial x} (\frac{- 8 x + 5}{2 y ^{2} + 1})

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{2 y ^{2} + 1} \frac{\partial}{\partial x} (- 8 x + 5)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{2 y ^{2} + 1} (- \frac{\partial}{\partial x} (8 x) + \frac{\partial}{\partial x} (5))

\frac{\partial}{\partial x} (8 x) = 8

\frac{\partial}{\partial x} (5) = 0

= \frac{1}{2 y ^{2} + 1} (- 8 + 0)

Simplify

\frac{1}{2 y ^{2} + 1} (- 8 + 0) : - \frac{8}{2 y ^{2} + 1}

= - \frac{8}{2 y ^{2} + 1}

y^{^{'}} + \frac{1}{x} y = 7 x^{2}

\frac{d}{d x} (ln (x) - \frac{1}{x})

x = 0 \sum \infty sin (x)

\int x x - 2 d x

\frac{e ^{x}}{cos ( x )}

What is (\partial)/(\partial x)((-8x+5)/(2y^2+1)) ?
The answer to (\partial)/(\partial x)((-8x+5)/(2y^2+1)) is -8/(2y^2+1)