Derivative

taurus · #1 $Old$ September 18th, 2007, 03:43 PM

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?

Krizalid · #2 $Old$ September 18th, 2007, 03:48 PM

You better use

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

Then take the derivative.

AfterShock · #3 $Old$ September 18th, 2007, 04:02 PM

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

$\frac{5x+8}{8x+3}\cdot \frac{49}{(5x+8)^2} = \frac{49}{(5x+8)(8x+3)}$

taurus · #4 $Old$ September 18th, 2007, 04:03 PM

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?)