# Thread: Chain rule and Sec^2 problem

1. ## Chain rule and Sec^2 problem

Hi there, i am having problems evaluating
$
F(x)=tan (x^2)$

which is differentiated to...
$
2x sec^2 (x^2)
$

when
$
x=\pi /4, for f'
$

when i put the numbers in i get the wrong answer, can someone explain what i am doing wrong, also im not sure what to do with the sec^2, how do you work this out on a calculator (for future reference) ???

Many thanks

2. Throw that calculator out
Go to this web site

3. How do make sec^2 in terms of cos??? whats the formula?

4. That is not the problem that you posted.
Please post only the exact question.

The derivative of $\tan(x^2)$ is $2x\sec^2(x^2)$.

Why would you change anything about that. $\sec(\frac{\pi}{4})=\sqrt2$.

5. how is sec $\sec(\frac{\pi}{4})=\sqrt2$ ??? not sure how its the $\sqrt2$

6. With all do respect, I must tell you some simple facts.
If you do not know the very basic relationships among the trig functions, there is absolutely no reason for you to even bother with these problems. We can of course tell you the answer. But what good does that do for you?

If you do not understand why $\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$ implies that $\sec(\frac{\pi}{4})=\sqrt2$ then you are wasting your own time.

7. Originally Posted by Plato
With all do respect, I must tell you some simple facts.
If you do not know the very basic relationships among the trig functions, there is absolutely no reason for you to even bother with these problems. We can of course tell you the answer. But what good does that do for you?

If you do not understand why $\cos(\frac{\pi}{4})=\frac{\sqrt2}{2}$ implies that $\sec(\frac{\pi}{4})=\sqrt2$ then you are wasting your own time.
We all forget things from time to time...

Madmax - you need to know that $\displaystyle \sec{x} = \frac{1}{\cos{x}}$.

8. yes, i remember $\displaystyle \sec{x} = \frac{1}{\cos{x}}$, however, how do you manipulate this to form $\frac {2}{cos(2x)+1}$ ??? is this some form of differentiation, or am i getting confused?

yes, i remember $\displaystyle \sec{x} = \frac{1}{\cos{x}}$, however, how do you manipulate this to form $\frac {2}{cos(2x)+1}$ ??? is this some form of differentiation, or am i getting confused?
What has this got to do with the question you originally posted? It is not related at all as far as I can see.

how is sec $\sec(\frac{\pi}{4})=\sqrt2$ ??? not sure how its the $\sqrt2$
Since $\frac{\pi}{4}+ \frac{\pi}{4}= \frac{\pi}{2}$, $\frac{\pi}{2}$ is its own conjugate and $sin(\pi/4)= cos(\pi/4)$. Since, for any angle $\theta$, $sin^2(\theta)+ cos^2(\theta)= 1$, $sin^2(\pi/4)+ cos^2(\pi/4)= sin^2(\pi/4)+ sin^2(\pi/4)= 2sin^2(\pi/4)= 1$. From that $sin^2(\pi/4)= \frac{1}{2}$ so $cos(\pi/4)= sin(\pi/4)= \sqrt{\frac{1}{2}}= \frac{1}{\sqrt{2}}$. From that,
$sec(\pi/4)= \frac{1}{\frac{1}{\sqrt{2}}}= \sqrt{2}$.

Or: because $\frac{\pi}{4}+ \frac{\pi}{4}= \frac{\pi}{2}$, there exist a right triangle with both acute angles equal to $\frac{\pi}{4}$. Because the two angles are the same, so are the legs. Take them to have length "1". Then, by the Pythagorean theorem, the length of the hypotenuse is given by $c^2= 1^2+ 1^2= 2$ so that $c= \sqrt{2}$. $cos(\theta)$= "near side over hypotenuse"= [/tex]\frac{1}{\sqrt{2}}[/tex] so that $sec(\pi/4)= \frac{1}{\frac{1}{\sqrt{2}}}= \sqrt{2}$.

To do that on a calculator, first make sure you calculator is in radian mode, then enter "pi" "/" "4", press the "cosine" key, then the reciprocal, "1/x" key.

yes, i remember $\displaystyle \sec{x} = \frac{1}{\cos{x}}$, however, how do you manipulate this to form $\frac {2}{cos(2x)+1}$ ??? is this some form of differentiation, or am i getting confused?
You don't! They are NOT equal (for example, at $x= \pi/2$, cos(x)= 0 so sec(0) does not exist but $\frac{2}{cos(2(0))+ 1}= 2$.) and no one here has suggested they are equal. Where did you get that formula?
You don't! They are NOT equal (for example, at $x= \pi/2$, cos(x)= 0 so sec(0) does not exist but $\frac{2}{cos(2(0))+ 1}= 2$.) and no one here has suggested they are equal. Where did you get that formula?