#Can someone explain the derivative of 1/3ln7x-1

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stoic pewterBOT
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sour pendant
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dy/dx = 1/3 * 1/7x-1 * 7

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doesnt that simplify to 1/3x-1?

sour pendant
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every online calculator says its 1/3x tho

old coyote
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Parentheses.

sour pendant
old coyote
# sour pendant wdym

I mean... add parentheses... so that I know which of the like fifty possible different functions you're actually talking about.

sour pendant
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(1/3)(ln7x-1)

old coyote
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Okay.

sour pendant
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or (ln7x-1)/3

old coyote
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Then 1/3 is a constant multiple.

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d/dx ln(kx) = 1/kx * k = 1/x.

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A way to think of it is with log properties: ln(kx) = ln(k) + ln(x)

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And ln(k) is a constant, whose derivative is 0.

sour pendant
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k is -1 here right

old coyote
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...no. k is 7.

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You did mean to write ln(7x), right?

sour pendant
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ln(7x-1)

old coyote
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This is why parentheses are important.

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d/dx ln(7x - 1)/3
= 1/3 * 1/(7x - 1) * 7
= 7/(21x - 3)```
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,w d/dx ln(7x - 1)/3

odd rampartBOT
old coyote
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See?

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You were getting the wrong answer because you were asking the wrong question.

sour pendant
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oh sorry i thought 1/3 * ln(7x-1) was the same as (ln(7x-1))/3

old coyote
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Remember order of operations?

old coyote
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...what was even the goal here?

sour pendant
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find derivative of the function by applying ln

old coyote
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...but why tho? Exponent rule + chain rule + quotient rule.

sour pendant
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yeah id rather do that but this is on my test

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i hate logs cause i didnt take pre cal