# Derivatives of inverse trig functions.

Printable View

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Nov 22nd 2010, 06:55 PM
FullBox
Derivatives 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 PM
dwsmith
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 PM
FullBox
Is this better for the first two problems?

http://oi55.tinypic.com/muj5ht.jpg
• Nov 22nd 2010, 07:18 PM
dwsmith
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 PM
FullBox
But isn't it -4sin^4(x)?

http://oi55.tinypic.com/2pyxdme.jpg
• Nov 22nd 2010, 07:37 PM
dwsmith
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 PM
FullBox
I'm sorry I don't understand. I thought I had already taken the derivative at that step.
• Nov 22nd 2010, 07:48 PM
dwsmith
$\displaystyle \displaystyle \frac{1}{6cos^4x}*24*cos^3(x)*(-sin(x))$ 1st derivative of ln, then 6cos^4(x), and then (cos(x)).
• Nov 22nd 2010, 08:20 PM
TheCoffeeMachine
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 PM
FullBox
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 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
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.

The third problem is bit more involved. Did you follow the second way that I've found the derivative for the first one?
• Nov 22nd 2010, 08:58 PM
FullBox
Quote:

Originally Posted by TheCoffeeMachine
The third problem is bit more involved. Did you follow the second way that I've found the derivative for the first one?

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 PM
Soroban
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 PM
TheCoffeeMachine
Quote:

Originally Posted by FullBox
Yes (or at least I think so). I actually understood your first explanation a lot better than the second one with the logarithms.

The easiest way of doing the 3rd problem is the same as the second way I found the derivative of #1 (albeit
taking the logarithm of both sides), that's why I asked you. But Soroban did the whole thing, anyway, now.
• Nov 22nd 2010, 09:18 PM
FullBox
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).
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last