What is (\partial)/(\partial x)(1+c/x-b/(x^3)) ?

The answer to (\partial)/(\partial x)(1+c/x-b/(x^3)) is (3b)/(x^4)-c/(x^2)

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(\partial)/(\partial x)(1+c/x-b/(x^3))

\frac{\partial}{\partial x} (1 + \frac{c}{x} - \frac{b}{x ^{3}})

\frac{3 b}{x ^{4}} - \frac{c}{x ^{2}}

Solution steps

\frac{\partial}{\partial x} (1 + \frac{c}{x} - \frac{b}{x ^{3}})

Treat

c, b

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (1) + \frac{\partial}{\partial x} (\frac{c}{x}) - \frac{\partial}{\partial x} (\frac{b}{x ^{3}})

\frac{\partial}{\partial x} (1) = 0

\frac{\partial}{\partial x} (\frac{c}{x}) = - \frac{c}{x ^{2}}

\frac{\partial}{\partial x} (\frac{b}{x ^{3}}) = - \frac{3 b}{x ^{4}}

= 0 - \frac{c}{x ^{2}} - (- \frac{3 b}{x ^{4}})

Simplify

= \frac{3 b}{x ^{4}} - \frac{c}{x ^{2}}

k = 1 \sum \infty ln (1 + \frac{1}{k})

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

x \to 2 lim (arctan (\frac{( 2 x ^{2} - 8 )}{3 x ^{2} - 6 x}))

y = - x^{2} - 2 x, y = - 2 x - 4, x = - 3, x = 0

y^{^{'''}} + y^{^{'}} = tan (x)

What is (\partial)/(\partial x)(1+c/x-b/(x^3)) ?
The answer to (\partial)/(\partial x)(1+c/x-b/(x^3)) is (3b)/(x^4)-c/(x^2)