What is (\partial)/(\partial x)(ln(xz)+yln(x)-x^2+4) ?

The answer to (\partial)/(\partial x)(ln(xz)+yln(x)-x^2+4) is -2x+(y+1)/x

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(\partial)/(\partial x)(ln(xz)+yln(x)-x^2+4)

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

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

Solution steps

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

Treat

z, y

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{\partial}{\partial x} (4) = 0

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

Simplify

\frac{1}{x} + \frac{y}{x} - 2 x + 0 : - 2 x + \frac{y + 1}{x}

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

\int (x^{3} + 1) sin (x) d x

\int x (3 x - 2)^{2} d x

f (x) = \frac{x ^{2} + 3 x}{6}

\int (6 x^{6} - 6 x^{5} - 4 x^{3} + 2) d x

ln (cos (x)), 0

What is (\partial)/(\partial x)(ln(xz)+yln(x)-x^2+4) ?
The answer to (\partial)/(\partial x)(ln(xz)+yln(x)-x^2+4) is -2x+(y+1)/x