MHB Derivative of sec(tan(2/x)) - 65 Characters

[karush]

find the derivative of $\sec\left({\tan\left({\frac{2}{x}}\right)}\right)$

I couldn't seem to find a way to rewrite this to make it easier and the answer from the Ti looks pretty hefty

 

[Siron]

What is your result when you apply the chain rule?

 

[karush]

$$f'\left(g\left(x\right)\right)g'\left(x\right)$$
So if
$g=\tan\left({\frac{2}{x}}\right)$ and $f=\sec\left({g}\right)$
If OK now what...I don't need to simplify

 

[Jameson]

You have three functions really, don't forget $\frac{2}{x}$. What do you get when you apply the chain rule twice?

 

[karush]

Well if $h=\frac{2}{x}$ then $$f'\left(g'\left(h\right)h'\right)g'$$
I think anyway.

 

[Jameson]

Try evaluating this specific case instead of the general form. I want to see what you get for this problem when you take the derivative.

 

[karush]

Let me see if this correct
$$f'=\sec\left({g}\right)\tan\left({g}\right)
\d{g}{x}$$
$$g'=\sec^2\left({h}\right)\d{h}{x}$$
$$h'=-\frac{2}{x^2}\d{}{x}$$

 

[Jameson]

Yes that is looking good. So what would you say is the final answer for $\frac{d}{dx} \sec\left({\tan\left({\frac{2}{x}}\right)}\right)$? Not in pieces but altogether what is the final answer?

In general if we have $$f(g(h(x)))$$ the the derivative will be $$f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$$. That is messy to look at though so sometimes it's easier just to do the specific problem. :)

 

[karush]

$f'\left(g'\left(h\right)h'\right)g'$

$\sec\left({g}\right)\tan\left(g\right)\left[\sec^2\left({h}\right)\frac{-2}{x^2}\right]
\sec^2\left({h}\right)$

so if I subst in $g$ and $h$ is that it? kinda long answer?

 

[Jameson]

Sorry I missed this yesterday. :)

Can you write your answer explicitly, meaning no substitutions or writing in terms of $f,g,h$?

$f'\left(g'\left(h\right)h'\right)g'$ is not the correct general form of this type of calculation of the derivative. It would be $\displaystyle f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$.

The first step, starting with the outer most function would be this:

$$\sec \left[ \tan \left( \frac{2}{x} \right) \right]\tan \left[ \tan \left( \frac{2}{x} \right) \right]$$

Now go in one step and multiply the above by the derivative of $$\tan \left( \frac{2}{x} \right)$$.

Then go in one step more and multiply the above by the derivative of $$\frac{2}{x}$$.

I think you are very close and might see this but I'm not sure by your notation. What do you get as a final answer in full form? Yes it's going to be quite long. :)

 

[karush]

$$\sec \left[ \tan \left( \frac{2}{x} \right) \right]\tan \left[ \tan \left( \frac{2}{x} \right) \right]$$

Now go in one step and multiply the above by the derivative of $$\tan \left( \frac{2}{x} \right)$$.

$$\tan\left({\frac{2}{x}}\right)
\tan\left({\tan\left({\frac{2}{x}}\right)}\right)
\sec\left({\tan\left({\frac{2}{x}}\right)}\right)
$$

Then go in one step more and multiply the above by the derivative of $$\frac{2}{x}$$.

$$-\frac{2\tan\left({\frac{2}{x}}\right)
\tan\left({\tan\left({\frac{2}{x}}\right)}\right)
\sec\left({\tan\left({\frac{2}{x}}\right)}\right)}{x^2}$$

kinda hefty nice to have Quick Latex

 

[Jameson]

So, so close. The derivative of $\tan(x)$ is $\sec^2(x)$. Add an exponent to the first term in your numerator and that looks correct. :)

 

[karush]

do you mean this(Dull)

$$-\frac{2\sec^2\left({\frac{2}{x}}\right)
\tan\left({\tan\left({\frac{2}{x}}\right)}\right)
\sec\left({\tan\left({\frac{2}{x}}\right)}\right)}{x^2}$$

 

[Jameson]

Yep, that looks correct and Wolfram Alpha agrees too.

 

[karush]

thanks for help. I really got lost in this one.

My TI gave a (same) alternate answer, but wasn't to helpful since it didn't show any steps.

 

[MarkFL]

thanks for help. I really got lost in this one.

My TI gave a (same) alternate answer, but wasn't to helpful since it didn't show any steps.

Sometimes it can take more work than the original problem to determine if your result is equivalent to that spat out by a CAS. :D