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Help on derivatives

  • Thread starter hawkblader
  • Start date
  • #1

Homework Statement



Let h(t) = tan(3t+7)

Find h'(t) and h''(t)

I found h'(t) which is equal to 3(sec(3t+7))^2

But I can't seem to find h''(t)

How do I find the derivative of this? Could someone please teach me?

Is it a composition function? If it is, I think I see 4 functions.

Thanks in advance.
 

Answers and Replies

  • #2
Ibix
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There are at least two ways to do it. The key observation is that sec(x)=1/cos(x). That should give you the answer using standard tools for handling derivatives of functions.
 
  • #3
There are at least two ways to do it. The key observation is that sec(x)=1/cos(x). That should give you the answer using standard tools for handling derivatives of functions.
Yeah, I knew that sec(x) = 1/cos(x)

and I still couln't figure out what to do
 
  • #4
SteamKing
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Maybe if you substitute 1/cos for sec in h'(t) that will give you a clue.
 
  • #5
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Or you can differentiate sec(u) directly, using the formula d/dt(sec(u)) = sec(u)tan(u) * du/dt.
 
  • #6
Ibix
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You have already used what I would call the Chain Rule:
[tex]\frac{d}{dx}f_1(f_2(x))=f_1'(f_2(x))f_2'(x)[/tex]
to get h'(x). All you need to realise is that you can nest functions as deep as you like - just replace [itex]x[/itex] with [itex]f_3(x)[/itex] throughout and tack [itex]f_3'(x)[/itex] on the end:
[tex]\frac{d}{dx}f_1(f_2(f_3(x)))=f_1'(f_2(f_3(x)))f_2'(f_3(x))f_3(x)[/tex]
Then you need to work out what each of the [itex]f[/itex]s is here and dive in.
 

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