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How would I evaluate the following two derivatives.

$g(t)=\sin^2(x)+\cos^2(t)$

For this derivative I know $\sin^2+\cos^2(t)=1$

so there derivative of $1$ is $0$

For my second question I have to find the second derivative of $h(t)=\sec(3t)$ at $t=\pi$

I found the first derivative as $3\sec(3t)\tan(3t)$

The second derivative I got

$3[\sec(3t)(3\sec^2(3t))+\tan(3t)(3)\sec(3t)\tan(3t)]$

But how would I apply the $\pi$ to the $t$

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  • $\begingroup$ Note that $\sin(3\pi)=\sin(\pi)=0$, and $\cos(3\pi)=\cos(\pi)=-1$. Now the other trig functions are easy. Conveniently, $\tan$ dies. $\endgroup$ – André Nicolas Feb 16 '13 at 21:11
  • $\begingroup$ Where does the red one close in: $$3[\sec(3t)\color{red}{(}3\sec^2(3t)+\tan(3t)(3)\sec(3t)\tan(3t)]?$$ $\endgroup$ – Git Gud Feb 16 '13 at 21:15
  • $\begingroup$ Oh sorry at the end of 3t $\endgroup$ – Fernando Martinez Feb 16 '13 at 21:17
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We need to substitute $\pi$ for $t$ in our expression for the derivative.

So we will need $\tan(3\pi)$ and $\sec(3\pi)$. One is more likely to be comfortable with $\sin$ and $\cos$.

Note that $\sin(3\pi)=\sin(\pi)=0$, and $\cos(3\pi)=\cos(\pi)=-1$. Now the other trig functions are easy. Conveniently, $\tan$ dies at $3\pi$. You should end up with $-9$.

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  • $\begingroup$ yes I have gotten -9 too. $\endgroup$ – Fernando Martinez Feb 16 '13 at 21:32
  • $\begingroup$ because $3sec^2(3pi)$ is 3 $\endgroup$ – Fernando Martinez Feb 16 '13 at 21:33
  • $\begingroup$ Yes, basically because $\cos(3\pi)$, and therefore $\sec(3\pi)$, is $-1$. $\endgroup$ – André Nicolas Feb 17 '13 at 1:22

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