What is (\partial)/(\partial x)(8xln(xy-7)) ?

The answer to (\partial)/(\partial x)(8xln(xy-7)) is 8(ln(xy-7)+(xy)/(xy-7))

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(\partial)/(\partial x)(8xln(xy-7))

\frac{\partial}{\partial x} (8 x ln (x y - 7))

8 (ln (x y - 7) + \frac{x y}{x y - 7})

Solution steps

\frac{\partial}{\partial x} (8 x ln (x y - 7))

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{\partial}{\partial x} (x ln (x y - 7))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = ln (x y - 7)

= 8 (\frac{\partial}{\partial x} (x) ln (x y - 7) + \frac{\partial}{\partial x} (ln (x y - 7)) x)

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

\frac{\partial}{\partial x} (ln (x y - 7)) = \frac{y}{y x - 7}

= 8 (1 \cdot ln (x y - 7) + \frac{y}{y x - 7} x)

Simplify

8 (1 \cdot ln (x y - 7) + \frac{y}{y x - 7} x) : 8 (ln (x y - 7) + \frac{x y}{x y - 7})

= 8 (ln (x y - 7) + \frac{x y}{x y - 7})

\frac{d}{d x} (4 arccos (- 5 x))

1, 4 x, 0, 1

\frac{d}{d θ} (2 cos (θ) + 4 sin (θ))

\frac{d}{d x} ((sin (x))^{4} + sin (x^{4}))

\frac{1}{4 s + 2}

What is (\partial)/(\partial x)(8xln(xy-7)) ?
The answer to (\partial)/(\partial x)(8xln(xy-7)) is 8(ln(xy-7)+(xy)/(xy-7))