What is (\partial)/(\partial x)(9/x+3/y+xy) ?

The answer to (\partial)/(\partial x)(9/x+3/y+xy) is y-9/(x^2)

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(\partial)/(\partial x)(9/x+3/y+xy)

\frac{\partial}{\partial x} (\frac{9}{x} + \frac{3}{y} + x y)

y - \frac{9}{x ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{9}{x} + \frac{3}{y} + x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (\frac{9}{x}) + \frac{\partial}{\partial x} (\frac{3}{y}) + \frac{\partial}{\partial x} (x y)

\frac{\partial}{\partial x} (\frac{9}{x}) = - \frac{9}{x ^{2}}

\frac{\partial}{\partial x} (\frac{3}{y}) = 0

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

= - \frac{9}{x ^{2}} + 0 + y

Simplify

= y - \frac{9}{x ^{2}}

\frac{d}{d x} (x^{a r c t a n (x)})

\frac{d y}{d x} + 6 (tan (6 x)) y = 9 cos (6 x)

\frac{d}{d x} (8 x^{2} - x^{3})

\int_{0}^{x} cos (t) d t

What is (\partial)/(\partial x)(9/x+3/y+xy) ?
The answer to (\partial)/(\partial x)(9/x+3/y+xy) is y-9/(x^2)