MHB Trig derivatives general question

[bmanmcfly]

Hi, here I come again now with a problem, this time is more because the boo didn't really explain much deeper than the minimum.
This time it's not homework related ... Yet. Lol.

I just wanted to know how I would separate an equation like
[math]y=sin^2x cos3x[/math]

To find the derivativ;
I guess how would I split this up into the chain and product rules / power rules.

I unfortunately forget most of the trig, but without knowing, but while I'm keeping up with that, I need to make sure that I'm at least setting up the problems correctly.

Thanks.

 

[MarkFL]

I would use the product rule, and in the application of that, the power and chain rules as well, so your approach sounds good to me! :D

If you get stuck in this process, feel free to post where you get stuck and one of us will be glad to help.

edit: Just a $\LaTeX$ and general tip...precede trig (as well as log and other) pre-defined functions with a backslash and enclose arguments in parentheses, i.e.:

y=\sin^2(x)\cos(3x) gives you $$y=\sin^2(x)\cos(3x)$$

The parentheses make it absolutley clear what the arguments are, and the backslash writes the function name without being italicized to differentiate it from a string of variables.

 

[bmanmcfly]

I would use the product rule, and in the application of that, the power and chain rules as well, so your approach sounds good to me! :D

If you get stuck in this process, feel free to post where you get stuck and one of us will be glad to help.

edit: Just a $\LaTeX$ and general tip...precede trig (as well as log and other) pre-defined functions with a backslash and enclose arguments in parentheses, i.e.:

y=\sin^2(x)\cos(3x) gives you $$y=\sin^2(x)\cos(3x)$$

The parentheses make it absolutley clear what the arguments are, and the backslash writes the function name without being italicized to differentiate it from a string of variables.

Fair enough; for this example [math] 2\sin(x)cos(x)\cos(3x)+\sin^2(x)(-\sin(3x)(3))[/math]
Is this a good start?

 

[MarkFL]

Yes, looks good to me. You have observed all of the rules you mentioned and correctly applied them. (Yes)

 

[bmanmcfly]

Yes, looks good to me. You have observed all of the rules you mentioned and correctly applied them. (Yes)

Unfortunately, that was the relatively easy part... Albeit oddly confusing on its own, the simplification part is more difficult since I don't really remember much of the trig equivalences. :(

 

[MarkFL]

I would begin by factoring...do you see any factors common to both terms?

 

[bmanmcfly]

I would begin by factoring...do you see any factors common to both terms?

Ok, factoring works, but I was concerned about issues like where [math]\sin(2x)=2\sin(x)\cos(x)[/math] or other similar equivalences, but there was nothing relevant and for not I can't really find others that matter...

Starting to think I'm panicking over nothing... Yet.

I would mark this as solved, but I will return to this thread if I need further clarification of concepts... The book I'm using is good, but really made for a classroom setting, so it seems to skimp on the details.

 

[MarkFL]

You could factor the $\sin(x)$ out:

$$y'=\sin(x)(2\cos(x)\cos(3x)-3\sin(x)\sin(3x))$$

and then apply product-to-sum identities as follows:

$$y'=\sin(x)\left(2\left(\frac{\cos(2x)+\cos(4x)}{2} \right)-3\left(\frac{\cos(2x)-\cos(4x)}{2} \right) \right)$$

Factor $\frac{1}{2}$ out:

$$y'=\frac{1}{2}\sin(x)\left(2\left(\cos(2x)+\cos(4x) \right)-3\left(\cos(2x)-\cos(4x) \right) \right)$$

Distribute:

$$y'=\frac{1}{2}\sin(x)\left(2\cos(2x)+2\cos(4x)-3\cos(2x)+3\cos(4x) \right)$$

Collect like terms:

$$y'=\frac{1}{2}\sin(x)\left(5\cos(4x)-\cos(2x) \right)$$

 

[bmanmcfly]

You could factor the $\sin(x)$ out:

$$y'=\sin(x)(2\cos(x)\cos(3x)-3\sin(x)\sin(3x))$$

and then apply product-to-sum identities as follows:

$$y'=\sin(x)\left(2\left(\frac{\cos(2x)+\cos(4x)}{2} \right)-3\left(\frac{\cos(2x)-\cos(4x)}{2} \right) \right)$$

Factor $\frac{1}{2}$ out:

$$y'=\frac{1}{2}\sin(x)\left(2\left(\cos(2x)+\cos(4x) \right)-3\left(\cos(2x)-\cos(4x) \right) \right)$$

Distribute:

$$y'=\frac{1}{2}\sin(x)\left(2\cos(2x)+2\cos(4x)-3\cos(2x)+3\cos(4x) \right)$$

Collect like terms:

$$y'=\frac{1}{2}\sin(x)\left(5\cos(4x)-\cos(2x) \right)$$

Youve Figuratively broken my brain for the night.

I hadn't even considered factoring to that extent...

Thanks for the help.

 

[Prove It]

Hi, here I come again now with a problem, this time is more because the boo didn't really explain much deeper than the minimum.
This time it's not homework related ... Yet. Lol.

I just wanted to know how I would separate an equation like
[math]y=sin^2x cos3x[/math]

To find the derivativ;
I guess how would I split this up into the chain and product rules / power rules.

I unfortunately forget most of the trig, but without knowing, but while I'm keeping up with that, I need to make sure that I'm at least setting up the problems correctly.

Thanks.

Finding the derivative of this would be easier to first use a double angle formula to simplify, so that you don't need to use the chain rule as well as the product rule.

[math]\displaystyle \begin{align*} y &= \sin^2{(x)}\cos{(3x)} \\ y &= \frac{1}{2}\left[ 1 - \cos{(2x)} \right] \cos{(3x)} \\ \frac{dy}{dx} &= \frac{1}{2} \left\{ 2\sin{(2x)} \cos{(3x)} - 3\sin{(3x)}\left[ 1 - \cos{(2x)} \right] \right\} \end{align*}[/math]

 

[bmanmcfly]

Nevermind I figured that one out

 

[Jameson]

This is a bit of a dumb question, but if the answer I give is in radians, but the problem uses degrees, should I convert the degrees to radians before solving?

If you have to plug in angle values that are given to you in degree form then yes you should convert to radians. In this problem and almost all calculus trig problems the rules for differentiating sin(x), cos(x) and all of the others all are for an angle in radians. If you use degrees then the derivatives change. So use radians to find the actual values of sin(a) for angle a. The output of trig functions is a ratio so this part doesn't change with degrees or radians, just the angle you input changes.

 

[bmanmcfly]

If you have to plug in angle values that are given to you in degree form then yes you should convert to radians. In this problem and almost all calculus trig problems the rules for differentiating sin(x), cos(x) and all of the others all are for an angle in radians. If you use degrees then the derivatives change. So use radians to find the actual values of sin(a) for angle a. The output of trig functions is a ratio so this part doesn't change with degrees or radians, just the angle you input changes.

I figured it out by plugging the numbers into a triangle and the result was way off...

But that raises a question, how do the derivatives change with degrees rather than radians? Although I might come across the answer in the book...

Thanks for the help.

 

[SuperSonic4]

I figured it out by plugging the numbers into a triangle and the result was way off...

But that raises a question, how do the derivatives change with degrees rather than radians? Although I might come across the answer in the book...

Thanks for the help.

It is because you would have to use the chain rule when differentiating. As you probably know 1 radian is equal to $\frac{180}{\pi}^o$. Thus if you wanted to find $\dfrac{d}{dx} \sin(x^o)$ you'd have to write it as $\dfrac{d}{dx} \sin \left(\frac{180}{\pi} \cdot x \right)$ which, by the chain rule would be the same as $\frac{d}{dx} \sin \left(\frac{180}{\pi} \cdot x \right) \times \dfrac{d}{dx} \frac{180}{\pi} \cdot x)$

I believe the reason the argument must be in radians is related to the small angle approximation but I'm not 100% sure