What is (\partial)/(\partial y)(1-3/y+x) ?

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

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(\partial)/(\partial y)(1-3/y+x)

\frac{\partial}{\partial y} (1 - \frac{3}{y} + x)

\frac{3}{y ^{2}}

Solution steps

\frac{\partial}{\partial y} (1 - \frac{3}{y} + x)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (1) - \frac{\partial}{\partial y} (\frac{3}{y}) + \frac{\partial}{\partial y} (x)

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

\frac{\partial}{\partial y} (\frac{3}{y}) = - \frac{3}{y ^{2}}

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

= 0 - (- \frac{3}{y ^{2}}) + 0

Simplify

= \frac{3}{y ^{2}}

f (x) = (x + 1)^{x + 1}

y = sin (9 x) + cos (7 x)

\frac{d}{d x} (2 - 5 x^{2})

\int_{2}^{3} \frac{1}{3 x - 2} d x

f (x) = x^{e}

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