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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

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

x \to \infty lim (\frac{ln ( x )}{3 x})

\frac{1}{x - 1}

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x + 1 y)

f (x) = 5 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 4

\frac{d y}{d x}, y^{3} + y = x

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