Second derivative questions involving trig functions.

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strongarmj

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I have the problem y=2 tanx and we are requested to find y''
y'=2sec^2(x)
y''=4sec(x)tan(x)sec(x) if I have this correct, how do I simplify to the final form. The professor has only provided the derivative formulas for trig functions and no example exercises.

Any help would be greatly appreciated. Thanks in advance.
 
strongarmj said:
I have the problem y=2 tanx and we are requested to find y''
y'=2sec^2(x)
y''=4sec(x)tan(x)sec(x) if I have this correct, how do I simplify to the final form. The professor has only provided the derivative formulas for trig functions and no example exercises.

Any help would be greatly appreciated. Thanks in advance.
Click to expand...
As far as I can see you have calculated the y' and y" correctly.

But you have not indicated any desired form!
 
strongarmj said:
I have the problem y=2 tanx and we are requested to find y''
y'=2sec^2(x)
y''=4sec(x)tan(x)sec(x) if I have this correct, how do I simplify to the final form. The professor has only provided the derivative formulas for trig functions and no example exercises.

Any help would be greatly appreciated. Thanks in advance.
Click to expand...
I generally put things in terms of sines and cosines unless I've got a good reason not to. So I'd say
[math]y''(x) = 4 \dfrac{sin(x)}{cos^3(x)}[/math]
-Dan
 
strongarmj said:
I have the problem y=2 tanx and we are requested to find y''
y'=2sec^2(x)
y''=4sec(x)tan(x)sec(x) if I have this correct, how do I simplify to the final form. The professor has only provided the derivative formulas for trig functions and no example exercises.

Any help would be greatly appreciated. Thanks in advance.
Click to expand...
I would at least combine the secants, giving y''=4sec^2(x)tan(x). Beyond this, it's a matter of taste. Unless you have been given some class rules for simplifying, you can't be called wrong leaving it at that. Trig expressions can commonly be written in multiple equivalent ways, none of which is necessarily simpler than others, and therefore none is better.
 
I would say that your answer is not complete. As Dr Peterson said, combine sec(x)*sec(x) to sec2(x).
There are many alternate solutions but your answer is just fine after making the suggested change.
 
Let me do you a suggestion that might work on another occasion: Start applying the derivative rules since the definition of the function. Knowing that

\(\displaystyle f(x)=tan(x)=\frac {sin(x)}{cos(x)}\)

Then

\(\displaystyle f'(x)=\frac {cos^2(x)+sin^2(x)}{cos^2(x)}=\frac {1}{cos^2(x)}\)

Moreover

\(\displaystyle f''(x)=\frac {0-2cos(x)(-sin(x))}{cos^4(x)}=\frac {2cos(x)sin(x)}{cos^4(x)}=\frac {2sin(x)}{cos^3(x)}\)
 
Last edited:
strongarmj said:
I have the problem y=2 tanx and we are requested to find y''
y'=2sec^2(x)
y''=4sec(x)tan(x)sec(x) if I have this correct, how do I simplify to the final form. The professor has only provided the derivative formulas for trig functions and no example exercises.

Any help would be greatly appreciated. Thanks in advance.
Click to expand...

you could change \(\displaystyle \sec^2{x} \text{ to } (1+\tan^2{x})\) if you desire the second derivative in terms of a single trig function
 
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