Once again, another few problems I got stuck on in my practice set. Here is my work typed in Microsoft Word.
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.
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}$
$\displaystyle \displaystyle \therefore ~ \frac{dy}{dx} = \frac{4\left(\cos{\theta}\right)'}{\cos{\theta}} = -\frac{4\sin{\theta}}{\cos{\theta}} = -4\tan{\theta}.$
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.
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.
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.)
Hello, FullBox!
For the third problem, use Logarithmic Differentiation.
$\displaystyle y \:=\:(\sin5x)^{3x}$
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] $
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).