Find the second derivative of h(t)=tan(3t+7)

[hawkblader]

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.

 

[Ibix]

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.

 

[hawkblader]

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

 

[SteamKing]

Maybe if you substitute 1/cos for sec in h'(t) that will give you a clue.

 

[Mark44]

Or you can differentiate sec(u) directly, using the formula d/dt(sec(u)) = sec(u)tan(u) * du/dt.

 

[Ibix]

You have already used what I would call the Chain Rule:
$$\frac{d}{dx}f_1(f_2(x))=f_1'(f_2(x))f_2'(x)$$
to get h'(x). All you need to realize is that you can nest functions as deep as you like - just replace $x$ with $f_3(x)$ throughout and tack $f_3'(x)$ on the end:
$$\frac{d}{dx}f_1(f_2(f_3(x)))=f_1'(f_2(f_3(x)))f_2'(f_3(x))f_3(x)$$
Then you need to work out what each of the $f$s is here and dive in.