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 $\displaystyle \displaystyle \frac{cos^3(x)}{cos^4(x)}=\frac{1}{cos(x)}\rightar row \frac{-4sin(x)}{cos(x)}\rightarrow -4tan(x)$

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
$\displaystyle \displaystyle \frac{1}{6cos^4x}*24*cos^3(x)*(-sin(x))$ 1st derivative of ln, then 6cos^4(x), and then (cos(x)).

- Nov 22nd 2010, 08:20 PMTheCoffeeMachine
Fullbox, $\displaystyle \frac{d}{d\theta}\left(6\cos^4{\theta\right) \ne 24\cos^3{\theta}\cdot \frac{d}{d\theta}\left(\cos^4{\theta}\right)$. Don't treat it as if the power

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

$\displaystyle \displaystyle \frac{d}{d\theta}\left[6(\cos{\theta)^4\right] = 24(\cos{\theta})^3\cdot \frac{d}{d\theta}\left(\cos{\theta}\right) = -24\cos^3{\theta}\sin{\theta}$.

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

logarithmic function $\displaystyle \log{a^b} = b\log{a}$ and $\displaystyle \log{xy} = \log{x}+\log{y}$, we have:

$\displaystyle \ln\left[6(\cos{\theta})^4\right] = \ln{6}+\ln\left(\cos{\theta}\right)^4 = \ln{6}+4\ln\(\cos{\theta}$

$\displaystyle \displaystyle \therefore ~ \frac{dy}{dx} = \frac{4\left(\cos{\theta}\right)'}{\cos{\theta}} = -\frac{4\sin{\theta}}{\cos{\theta}} = -4\tan{\theta}.$ - 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:

$\displaystyle y \:=\:(\sin5x)^{3x}$

Take logs: .$\displaystyle \ln(y) \:=\:\ln(\sin5x)^{3x} \:=\:3x\cdot\ln(\sin 5x)$

Differentiate implicitly:

. . $\displaystyle \displaystyle \frac{1}{y}\cdot\frac {dy}{dx} \;=\;3x\cdot\frac{1}{\sin5x}\cdot\cos5x\cdot 5 + 3\cdot\ln(\sin5x) $

. . $\displaystyle \displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\frac{15x\cos5x}{\sin5x} + 3\ln(\sin5x) $

. . $\displaystyle \displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;15x\cot5x + 3\ln(\sin5x)$

. . . . $\displaystyle \displaystyle \frac{dy}{dx} \;=\;y\cdot[15x\cot5x + 3\ln(\sin 5x)] $

. . . . $\displaystyle \displaystyle \frac{dy}{dx} \;=\;(\sin5x)^{3x}\left[15x\cot5x + 3\ln(\sin5x)\right] $

- 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).