# Thread: Even More Natural Log Problems With trig =D

1. ## Even More Natural Log Problems With trig =D

Ok this is the problem
$\displaystyle \ln|sec5x + cot4x|$

I assumed that the ln Can be distributed so i converted it to

$\displaystyle \ln sec5x + \ln cot(4x)$

and thus

$\displaystyle \frac{dy}{dx} = \frac{sec 5x tan 5x}{sec 5x} + \frac {-csc^2 5x}{cot 4x}$

which then eventually reduced down to

$\displaystyle \frac{(tan 5x cot 4x) - csc^2 5x}{cot 4x}$

does this look correct?

2. Originally Posted by x5pyd3rx
Ok this is the problem
$\displaystyle \ln|sec5x + cot4x|$

I assumed that the ln Can be distributed so i converted it to

$\displaystyle \ln sec5x + \ln cot(4x)$

and thus

$\displaystyle \frac{dy}{dx} = \frac{sec 5x tan 5x}{sec 5x} + \frac {-csc^2 5x}{cot 4x}$

which then eventually reduced down to

$\displaystyle \frac{(tan 5x cot 4x) - csc^2 5x}{cot 4x}$

does this look correct?
No No No No .
In general: $\displaystyle ln(a+b) \neq ln(a) + ln(b)$.

and what happened for the absolute values ?

3. OK so the absolute value signs mean i would have to do 2 equations one for negative and one for positive ?

4. Originally Posted by x5pyd3rx
Ok this is the problem
$\displaystyle \ln|sec5x + cot4x|$

I assumed that the ln Can be distributed so i converted it to

$\displaystyle \ln sec5x + \ln cot(4x)$

and thus

$\displaystyle \frac{dy}{dx} = \frac{sec 5x tan 5x}{sec 5x} + \frac {-csc^2 5x}{cot 4x}$

which then eventually reduced down to

$\displaystyle \frac{(tan 5x cot 4x) - csc^2 5x}{cot 4x}$

does this look correct?
ouch...as the beatles song goes, "because i told you before, ooo, you can't do that."

anyway, define
$\displaystyle f(x)= \ln|x|, g(x)=(\sec5x + \cot4x)$

then apply chain rule and note that there is yet another small application of the chain rule embedded within $\displaystyle g(x).$ So you'll 2 chain rules to perform to get your answer.

5. Im Completly lost.

6. Wait so your saying that i should calculate the derivative for g(x) for all values greater than 0 and then less than 0 and then back into f(x)?

7. chain rule: $\displaystyle f(g(x))^{'} = f^{'}(g(x))g^{'}(x)$

deriv of $\displaystyle ln|x|$ evaluated at $\displaystyle (sec5x + cot4x)$ times the deriv of $\displaystyle (sec5x + cot4x)$, where for example the deriv of $\displaystyle \sec(5x)$ is $\displaystyle 5\tan(5 x)\sec(5 x)$.

8. Excuse the triple posting but, i found something stating that

the derivative of a absolute value is u/|u| *du/dx if i can assume that then my answer would be simple

$\displaystyle \ln|sec5x+cot4x|$
$\displaystyle \frac{d}{dx} = \frac{\frac{sec5x + cot4x * sec5x tan5x - csc^2x}{|sec5x +cot4x|}}{|sec5x + cot4x|}$

This is possible because the derivative of ln states that du/dx = u'/u
so then i would simply just cancel the denomerators of both fractions. and my answer would be the numerator's Numerator.
$\displaystyle frac{sec5x + cot4x * sec5x tan5x - csc^2x}{|sec5x +cot4x|}$

9. Yes our answers would be the same. Except i forgot to "derive properly" thank you for your help

10. Originally Posted by x5pyd3rx
Excuse the triple posting but, i found something stating that

the derivative of a absolute value is u/|u| *du/dx if i can assume that then my answer would be simple

$\displaystyle \ln|sec5x+cot4x|$
$\displaystyle \frac{d}{dx} = \frac{\frac{sec5x + cot4x * sec5x tan5x - csc^2x}{|sec5x +cot4x|}}{|sec5x + cot4x|}$

This is possible because the derivative of ln states that du/dx = u'/u
so then i would simply just cancel the denomerators of both fractions. and my answer would be the numerator's Numerator.
$\displaystyle frac{sec5x + cot4x * sec5x tan5x - csc^2x}{|sec5x +cot4x|}$
the absolute value appears because the principal branch of the natural logarithm is chosen so as to restrict the domain of the function to positive reals. There is no such thing as a formula for taking the derivative of an "absolute value." First of all, the absolute value, of say $\displaystyle x$, does not have a proper limit...thus no derivative...but anyway, the bottom line is that with the substitutions i show above, this should be a piece of cake, assuming you've learned the chain rule. I really don't see what's hanging you up.

11. Originally Posted by x5pyd3rx
Yes our answers would be the same. Except i forgot to "derive properly" thank you for your help

oh good, you got rid of that line about formulas for taking derivs of absolute values...