Differentiation (1 Viewer)

[xibu34]

Can anybody help me differentiate

I understand it's the chain rule but I keep on making errors and subbing in my t value doesn't give the rate of change I want, so I just want to double check I have the right solution.

 

[Drongoski]

 

[Drongoski]

Let f(t) = 0.02t + 0.5cos(pi/6t - 4pi)

f'(t) = 0.02 - pi/12sin(pi/6t - 4pi)

u = 7500, u' = 0, v = e^(0.02t + 0.5cos(pi/6t - 4pi), v' = (0.02 - pi/12sin(pi/6t - 4pi) x e^(0.02t + 0.5cos(pi/6t - 4pi))

g'(t) = uv' + u'v (using product rule)

Therefore g'(t) = 7500(0.02 - pi/12sin(pi/6t - 4pi) x e^(0.02t + 0.5cos(pi/6t - 4pi))

I'm not sure if I made any mistakes, but the process should be right

Edit: Soz for the lack of Latex, also, you can just use the fact that if f(x) = e^g(x), f'(x) = g'(x) x e^g(x) as Drongoski has done instead of the product rule, but both works

7500 is only a simple constant multiple; so product rule not required. That is:

 

[xibu34]

Weird, I got the same answer but when I sub in my t value it doesn't give me the values it should, for example when I sub in t=29.854 the derivative should be 0 but I get something wack on my calculator.

 

[Drongoski]

Weird, I got the same answer but when I sub in my t value it doesn't give me the values it should, for example when I sub in t=29.854 the derivative should be 0 but I get something wack on my calculator.

Is calculator in degree or radian mode? Maybe this is irrelevant.

 

[xibu34]

Is calculator in degree or radian mode? Maybe this is irrelevant.

Radians, I'll have to check again but

Is calculator in degree or radian mode? Maybe this is irrelevant.

I got 0.05, close enough

 

[MathNerd!]

Can anybody help me differentiate
View attachment 38111
I understand it's the chain rule but I keep on making errors and subbing in my t value doesn't give the rate of change I want, so I just want to double check I have the right solution.

Chain rule is not necessary for this question. Functions of the form e^g(t) differentiate as g'(t).e^f(t). In other words, you can differentiate the exponent and put it out the front of g(t) ... much quicker!
BTW - You might also like to check out a new calculator approved for NSW HSC exams: TI-30XPlusMathPrint ... you can get it from OfficeWorks for around $37.00. Check out this sample video.

If you have to substitute a value for t as part of the question, you can do it all on the calculator by defining f(x) as the function and g(x) by first principles, set the step value small and voila!

 

[MJRey]

Where the heck is that question from? It looks terrifying.

 

[Average Boreduser]

Where the heck is that question from? It looks terrifying.

I thought this was 3U? its differentiating an exponential and trig function using product rule right?
note: idfk, I just learn how to diff an exponential so I'm just judging from that

 

[MJRey]

I mean from what textbook or paper did they get this from, cause I've never seen anything like it.