Once again, another few problems I got stuck on in my practice set. Here is my work typed in Microsoft Word.

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- Nov 22nd 2010, 06:55 PMFullBoxDerivatives of inverse trig functions.
Once again, another few problems I got stuck on in my practice set. Here is my work typed in Microsoft Word.

- Nov 22nd 2010, 07:01 PMdwsmith
1: just negative sine no 4 simplifies to -4tanx

2: 2x^5 is what you derive for the second part not 2x^10

3: The derivative I am getting with CAS is 3*(sin5x)^{3x-1}*(sin5x)*ln(sin5x)+5xcos(5x) but I can't obtain that by hand. - Nov 22nd 2010, 07:12 PMFullBox
Is this better for the first two problems?

http://oi55.tinypic.com/muj5ht.jpg - Nov 22nd 2010, 07:18 PMdwsmith
1 yes but

2 correct - Nov 22nd 2010, 07:28 PMFullBox
But isn't it -4sin^4(x)?

http://oi55.tinypic.com/2pyxdme.jpg - Nov 22nd 2010, 07:37 PMdwsmith
No you taking the derivative on the inner part of (cosx)^3 so that is the derivative of cosx which is negative sine

- Nov 22nd 2010, 07:45 PMFullBox
I'm sorry I don't understand. I thought I had already taken the derivative at that step.

- Nov 22nd 2010, 07:48 PMdwsmith
1st derivative of ln, then 6cos^4(x), and then (cos(x)).

- Nov 22nd 2010, 08:20 PMTheCoffeeMachine
Fullbox, . Don't treat it as if the power

was on theta, rather than being on outside. For clarity/simplicity, write it this way:

.

It might have been easier to note that, from the start, since by the properties of the

logarithmic function and , we have:

- Nov 22nd 2010, 08:41 PMFullBox
Oh okay thanks guys. I'm just horrible with syntax, that part really had me confused. I know you guys solved it already but I just want to update it onto the original work for the record's sake.

http://oi55.tinypic.com/4jt6cg.jpg

In regards to the third problem, I am still somewhat stuck. dwsmith, can you tell me what method you used in your post as opposed to typing out the answer? I want to try and figure it out myself I just don't know what to do. I'm unsure of what you did and in what order. - Nov 22nd 2010, 08:45 PMTheCoffeeMachine
- Nov 22nd 2010, 08:58 PMFullBox
Yes (or at least I think so). I actually understood your first explanation a lot better than the second one with the logarithms.

From what I understood, you first find the derivative of the entire function and then you multiplied it by the derivative of the inner-most part of the function. My mistake was that I was multiplying the derivative of the entire function by the derivative of the inner-most function*to the power of 4*. In reality, I was not actually computing the derivative of the innermost function since I was misunderstanding the syntax (the exponent being before or after the variable for a trigonometric function, like sin^(2)(x) or sin(x)^2.) - Nov 22nd 2010, 09:03 PMSoroban
Hello, FullBox!

For the third problem, use Logarithmic Differentiation.

Quote:

Take logs: .

Differentiate implicitly:

. .

. .

. .

. . . .

. . . .

- Nov 22nd 2010, 09:08 PMTheCoffeeMachine
- Nov 22nd 2010, 09:18 PMFullBox
I'm not so sure that I understand Soroban's way completely.

I understand the first step with the natural logs.

The next step I think I understand, but I'm not sure. I haven't reached the point where I can write down what I got without saying what I did in between so maybe that is what is confusing me.

From what it seems, we know ln(x) = 1/x. In this case, x=sin5x so he got 1/(sin5x). cos5x is the derivative of sin5x. 5 is the derivative of 5x. However, I am not so sure about what happens in the end of that line, where he Soroban writes 3*ln(sin5x).