# trig derivs.

• February 29th 2008, 09:21 AM
mathlete
trig derivs.
Find the derivative of the function below.
y(t) = sec^6(t + 1)

can i just get some help...this one just seems weird

thanks
• February 29th 2008, 09:53 AM
TKHunny
It's just a mega-chain rule deal. I'm not sure what help there is, other than demonstrating it. Once you see it, you'll have it firmly in your mind and never trip over another one.

$y'(t) = \left(6*[sec(t+1)]^{5}\right)*\left(sec(t+1)tan(t+1)\right)*(1)$

I added the final "(1)" only to emphasize that the argument of the secant also needed some attention.

I like to write these out in a thoretical sense. It helps build understanding of the phases of the solution.

If z(t) = h(g(r(t))), then z'(t) = h'(g(r(t)))*g'(r(t))*r'(t).

In your case, you have h(t) = t^6, g(t) = sec(t), and r(t) = t+1.
• February 29th 2008, 10:01 AM
mathlete
that helps alot, thanks. But i notice that i need the derivative of sec right, how would i get that?

Quote:

Originally Posted by TKHunny
It's just a mega-chain rule deal. I'm not sure what help there is, other than demonstrating it. Once you see it, you'll have it firmly in your mind and never trip over another one.

$y'(t) = \left(6*[sec(t+1)]^{5}\right)*\left(sec(t+1)tan(t+1)\right)*(1)$

I added the final "(1)" only to emphasize that the argument of the secant also needed some attention.

I like to write these out in a thoretical sense. It helps build understanding of the phases of the solution.

If z(t) = h(g(r(t))), then z'(t) = h'(g(r(t)))*g'(r(t))*r'(t).

In your case, you have h(t) = t^6, g(t) = sec(t), and r(t) = t+1.

• March 1st 2008, 05:06 AM
TKHunny
Well, you can derive it from fundamental principles if you like, or you can just memorize it. You should know these two:

$\frac{d}{dx}\tan(x)\;=\;\sec^{2}(x)$

$\frac{d}{dx}\sec(x)\;=\;\sec(x)\tan(x)$