What is the d/(dy)(2ln(3x+4y)) ?

The d/(dy)(2ln(3x+4y)) is 8/(3x+4y)

What is the first d/(dy)(2ln(3x+4y)) ?

The first d/(dy)(2ln(3x+4y)) is 8/(3x+4y)

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\frac{d}{d y} (2 ln (3 x + 4 y))

\frac{8}{3 x + 4 y}

Solution steps

\frac{d}{d y} (2 ln (3 x + 4 y))

Treat

x

as a constant

Take the constant out:

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

= 2 \frac{d}{d y} (ln (3 x + 4 y))

Apply the chain rule:

\frac{1}{3 x + 4 y} \frac{d}{d y} (3 x + 4 y)

= \frac{1}{3 x + 4 y} \frac{d}{d y} (3 x + 4 y)

\frac{d}{d y} (3 x + 4 y) = 4

= 2 \cdot \frac{1}{3 x + 4 y} \cdot 4

Simplify

2 \cdot \frac{1}{3 x + 4 y} \cdot 4 : \frac{8}{3 x + 4 y}

= \frac{8}{3 x + 4 y}

x \to 0 lim (\frac{4 cos ( x - 4 )}{x})

\int_{0}^{1} \frac{x + 1}{x + 3} d x

x \to - 8 lim (\frac{x ^{2} + 7}{x + 8})

(- 4, - 1), (3, 6)

\int x^{2} e^{x} d x

What is the d/(dy)(2ln(3x+4y)) ?
The d/(dy)(2ln(3x+4y)) is 8/(3x+4y)
What is the first d/(dy)(2ln(3x+4y)) ?
The first d/(dy)(2ln(3x+4y)) is 8/(3x+4y)