What is the derivative of 3ln(x+ln(y)) ?

The derivative of 3ln(x+ln(y)) is (3y+x\frac{dy)/(dx)}{xy}

What is the first derivative of 3ln(x+ln(y)) ?

The first derivative of 3ln(x+ln(y)) is (3y+x\frac{dy)/(dx)}{xy}

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\frac{d}{d x} (3 ln (x) + ln (y))

\frac{3 y + x \frac{d y}{d x}}{x y}

Solution steps

\frac{d}{d x} (3 ln (x) + ln (y))

Simplify

3 ln (x) + ln (y) : ln (x^{3} y)

= \frac{d}{d x} (ln (x^{3} y))

Apply the chain rule:

\frac{1}{x ^{3} y} \frac{d}{d x} (x^{3} y)

= \frac{1}{x ^{3} y} \frac{d}{d x} (x^{3} y)

\frac{d}{d x} (x^{3} y) = 3 x^{2} y + \frac{d y}{d x} x^{3}

= \frac{1}{x ^{3} y} (3 x^{2} y + \frac{d y}{d x} x^{3})

Simplify

\frac{1}{x ^{3} y} (3 x^{2} y + \frac{d y}{d x} x^{3}) : \frac{3 y + x \frac{d y}{d x}}{x y}

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

(1, 3), (- 2, - 3)

\frac{\partial}{\partial y} (e^{x^{2} + y})

\frac{d y}{d t} = 8 y, y (4) = 4

\frac{1}{2 s}

f (x) = cos (\frac{1}{x})

What is the derivative of 3ln(x+ln(y)) ?
The derivative of 3ln(x+ln(y)) is (3y+x\frac{dy)/(dx)}{xy}
What is the first derivative of 3ln(x+ln(y)) ?
The first derivative of 3ln(x+ln(y)) is (3y+x\frac{dy)/(dx)}{xy}