What is (\partial)/(\partial x)(4ln(xy)-(x+y)^2+5) ?

The answer to (\partial)/(\partial x)(4ln(xy)-(x+y)^2+5) is -2x+4/x-2y

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(\partial)/(\partial x)(4ln(xy)-(x+y)^2+5)

\frac{\partial}{\partial x} (4 ln (x y) - (x + y)^{2} + 5)

- 2 x + \frac{4}{x} - 2 y

Solution steps

\frac{\partial}{\partial x} (4 ln (x y) - (x + y)^{2} + 5)

Simplify

4 ln (x y) - (x + y)^{2} + 5 : 4 ln (x y) - x^{2} - 2 x y - y^{2} + 5

= \frac{\partial}{\partial x} (4 ln (x y) - x^{2} - 2 x y - y^{2} + 5)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4 ln (x y)) - \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (2 x y) - \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (5)

\frac{\partial}{\partial x} (4 ln (x y)) = \frac{4}{x}

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

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

\frac{\partial}{\partial x} (y^{2}) = 0

\frac{\partial}{\partial x} (5) = 0

= \frac{4}{x} - 2 x - 2 y - 0 + 0

Simplify

= - 2 x + \frac{4}{x} - 2 y

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

\int (x ln (x)) d x

\int (x + 1) y^{2} \cdot sin (\frac{x y}{3}) d x

x \to 0 - lim (g (t))

\frac{d}{d x} (ln (x y^{2}))

What is (\partial)/(\partial x)(4ln(xy)-(x+y)^2+5) ?
The answer to (\partial)/(\partial x)(4ln(xy)-(x+y)^2+5) is -2x+4/x-2y