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

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

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(\partial)/(\partial x)(1/x+1/(y+z))

\frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{y + z})

- \frac{1}{x ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{y + z})

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (\frac{1}{y + z}) = 0

= - \frac{1}{x ^{2}} + 0

Simplify

= - \frac{1}{x ^{2}}

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

\frac{d}{d x} ((3 x^{2} + 5 x + 1)^{\frac{3}{2}})

f (x) = lo g_{41} (2 x - 4)

\frac{d}{d x} (ln (x) cos (x))

x^{2} y^{^{''}} + 3 x y^{^{'}} - 15 y = 0

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