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  #1  
Old September 18th, 2007, 03:43 PM
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Question Derivative

I got the question f '(x) given:
f(x) = ln((8x+3)/(5x+8))

Now i first used quotient rule for inside bit and got:

(5x+8*8 - 8x+3*5) / (5x+8)^2

Now to use the chain rule for ln((8x+3)/(5x+8)) i am not sure, could someone help please?
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  #2  
Old September 18th, 2007, 03:48 PM
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You better use

\ln\frac ab=\ln a-\ln b, for suitably a,b

Then take the derivative.
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Old September 18th, 2007, 04:02 PM
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Quote:
Originally Posted by taurus View Post
I got the question f '(x) given:
f(x) = ln((8x+3)/(5x+8))

Now i first used quotient rule for inside bit and got:

(5x+8*8 - 8x+3*5) / (5x+8)^2

Now to use the chain rule for ln((8x+3)/(5x+8)) i am not sure, could someone help please?
There's a simple way to do it, but we'll do it your way.

If f(x) = \ln(\frac{8x + 3}{5x + 8}), then, f'(x) = \frac{1}{\frac{8x + 3}{5x + 8}} \cdot{derivative of inside here}

So, what is the derivative of the inside, that is, of:

\frac{8x + 3}{5x + 8}

Let's use the quotient rule.

\frac{8\cdot (5x + 8) - (8x + 3)\cdot 5}{(5x + 8)^2}

And thus, f'(x) = \frac{1}{\frac{8x + 3}{5x + 8}} \cdot \frac{8\cdot (5x + 8) - (8x + 3)\cdot 5}{(5x + 8)^2} = \frac{5x+8}{8x+3}\cdot \frac{8\cdot (5x + 8) - (8x + 3)\cdot 5}{(5x + 8)^2}

Note that 8\cdot (5x + 8) - (8x + 3)\cdot 5 = 49

\frac{5x+8}{8x+3}\cdot \frac{49}{(5x+8)^2} = \frac{49}{(5x+8)(8x+3)}
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Old September 18th, 2007, 04:03 PM
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alright yea i see, let me see if im right now:

--> ln(8x+3) - ln(5x+8)
--> 1/(8x+3) - 1/(5x+8)

(do i have to find the derivative of brackets, eg: 8x+3 => 8?)
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Old September 18th, 2007, 04:08 PM
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Uhhhmm, it's easier if we write f(x) as

f(x)=\ln(8x+3)-\ln(5x+8)\implies f'(x)=\frac8{8x+3}-\frac5{5x+8}
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Old September 18th, 2007, 04:12 PM
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Quote:
Originally Posted by Krizalid View Post
Uhhhmm, it's easier if we write f(x) as

f(x)=\ln(8x+3)-\ln(5x+8)\implies f'(x)=\frac8{8x+3}-\frac5{5x+8}
i thought lnx was 1/x?
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Old September 18th, 2007, 04:13 PM
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Quote:
Originally Posted by taurus View Post
i thought lnx was 1/x?
It is, but then you have to multiply by the "inside", and hence where the 5 and 8 come from.
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Old September 18th, 2007, 04:15 PM
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For example,

f(x) = \ln(5x) \Rightarrow f'(x) = \frac{1}{5x}\cdot 5 = \frac{1}{x}
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Old September 18th, 2007, 04:23 PM
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See The Chain Rule taurus.
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Old September 18th, 2007, 04:25 PM
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Quote:
Originally Posted by AfterShock View Post
For example,

f(x) = \ln(5x) \Rightarrow f'(x) = \frac{1}{5x}\cdot 5 = \frac{1}{x}
wait so shoudlnt that be 5/5x, like the other one?
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Old September 18th, 2007, 04:28 PM
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Quote:
Originally Posted by taurus View Post
wait so shoudlnt that be 5/5x, like the other one?
And what is \frac{5}{5x}? Do you see that it's equal to \frac{1}{x}?
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Old September 18th, 2007, 05:18 PM
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Quote:
Originally Posted by AfterShock View Post
And what is \frac{5}{5x}? Do you see that it's equal to \frac{1}{x}?
ohhhh yea
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Old September 19th, 2007, 03:51 PM
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I got another question:
Find the simplest function g(q) such that

g'(q) = sin( 5 q) +cos( 5 q)

im not sure how i should go about solving it?
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Old September 19th, 2007, 04:11 PM
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Well, you can integrate both sides to get g(q)
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Old September 19th, 2007, 04:18 PM
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what do you mean, how would i go about that?
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