derivatives of trig functions

Dorian Gray

Junior Member
Joined
Jan 20, 2012
Messages
143
Hello Mathematicians,

I have a math exercise that has me puzzled.

if H(theta)= (theta)sin(theta), what is H' and H''


I was able to find H', but I have no idea how to address H''. In my class, were were just learning the quotient rule, and we have not went over all of the trig derivatives yet. I have only learned sine, cos, -cos, and -sin. Therefore, any help with the second part would be greatly appreciated.

Here is a copy of my work.

Screen shot 2012-02-11 at 11.44.43 AM.jpg


Thank you for your time and knowledge
 
Last edited:
Hello Mathematicians,

I have a math exercise that has me puzzled.

if H(theta)= (theta)sin(theta), what is H' and H''
\
View attachment 1697
Dθ(θcos(θ)+sin(θ))=cos(θ)θsin(θ)+cos(θ)\displaystyle D_{\theta}(\theta\cos(\theta)+\sin(\theta))=cos(\theta)-\theta\sin(\theta)+cos(\theta)
 
thank you

Greetings Pka,

Thank you very much for your rapid response. Could you please tell me what you did (and why)? Essentially, how did you arrive to your answer?

Thanks again,
DG
 
Thank you very much for your rapid response. Could you please tell me what you did (and why)? Essentially, how did you arrive to your answer?
Sorry, but this is not a tutorial service.
 
Thank you for your help. 1. If you would've said"I just used theorem X" is all that I was looking for. and 2.I was simply trying to show a deeper interest than just using this site as a "I have a question, and all I care about is if somebody solves it correctly for me".
 
It's the first and second derivatives of H(θ)=θsinθ\displaystyle H(\theta)=\theta sin\theta

Use the product rule: d[fg]=fg+gf\displaystyle d[f\cdot g]=f\cdot g'+g\cdot f'

H(θ)=θcosθ+sinθ\displaystyle H'(\theta)=\theta \cdot cos\theta +sin\theta

Now, use the product rule again with this derivative:

θsinθ+1cosθ+cosθ\displaystyle \theta \cdot -sin\theta+1\cdot cos\theta + cos\theta

H(θ)=θsinθ+2cosθ\displaystyle H''(\theta)=-\theta sin\theta +2cos\theta
 
Thank You Galactus

Thank you very much Galactus. I cannot tell you how much I appreciate your extra details and explanation.
 
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