What is the d/(d{x)}(ln({x}+2{y}+3{z})) ?

The d/(d{x)}(ln({x}+2{y}+3{z})) is 1/({x)+2{y}+3{z}}

What is the first d/(d{x)}(ln({x}+2{y}+3{z})) ?

The first d/(d{x)}(ln({x}+2{y}+3{z})) is 1/({x)+2{y}+3{z}}

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

\frac{1}{x + 2 y + 3 z}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

= \frac{1}{x + 2 y + 3 z}

y^{^{'''}} = e^{- 0.2 x}

\frac{( s + 2 )}{( s ^{2} + 4 )}

(x^{2} + 4) (2 x - 6)

\frac{1}{( 1 + x ) ^{2}}

\int x^{5} cos (x^{2}) d x

What is the d/(d{x)}(ln({x}+2{y}+3{z})) ?
The d/(d{x)}(ln({x}+2{y}+3{z})) is 1/({x)+2{y}+3{z}}
What is the first d/(d{x)}(ln({x}+2{y}+3{z})) ?
The first d/(d{x)}(ln({x}+2{y}+3{z})) is 1/({x)+2{y}+3{z}}