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Second derivative of xtanx?

  • Thread starter escryan
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  • #1
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Homework Statement



Given y=xtanx, find y'' (second derivative)

Homework Equations



Uh... I'm not even sure if I'm using the right one....
d/dx(tanx) = sec^2x

The Attempt at a Solution



y=xtanx
y'= (x)(sec^2(x)) + (tanx)(1)
y'= xsec^2(x) + tanx

y'' = [(x)(2sec^3(x)) + sec^2(x)(1)] + sec^2x
y'' = 2xsec^3(x) + sec^2(x) + sec^2(x)

....

The key answer says it is supposed to be (2cosx + 2xsinx)/(cos^3(x))
 

Answers and Replies

  • #2
1,752
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you took the 2nd derivative wrong

[tex]y'=x\sec^{2}x+\tan x[/tex]

[tex]y''=x\cdot2\sec^{2-1}x\cdot\sec x\tan x+\sec^{2}x+\sec^{2}x[/tex]
[tex]=2x\sec^{2}x\tan x+2\sec^{2}x[/tex]

What you did was increase the power rather than decreasing it.

In general, the derivative of secant raised to a power is ... [tex]\frac{d}{dx}(\sec^{n}x)=n\sec^{n}x\tan x[/tex]
 
Last edited:
  • #3
13
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Oh wow, I'm an idiot..

So is the general rule d/dx(sec^n(x)) = nsec^n(x)tan(x) just a combination of the (x^n)' = nx^x-1 and d/dx(secx) = secxtanx?

What if I were to be givin d/dx(tan^n(x))... would the answer be like nsec^2(n-1)(x)? nsec^n(x)? ...

Thanks so much for your help, by the way :).
 
  • #4
1,752
1
No, don't forget the chain rule!!!

[tex]\frac{d}{dx}(\tan^{n}x)=n\tan^{n-1}x\sec^{2}x[/tex]
 
  • #5
13
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The chain rule?
... I've never actually seen that before.
Haha, I guess that explains a few things! I haven't been taught that yet.

Guess I'll go read up on that, and thanks again for your help! I really appreciate it :).
 
  • #6
1
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you took the 2nd derivative wrong

[tex]y'=x\sec^{2}x+\tan x[/tex]

[tex]y''=x\cdot2\sec^{2-1}x\cdot\sec x\tan x+\sec^{2}x+\sec^{2}x[/tex]
[tex]=2x\sec^{2}x\tan x+2\sec^{2}x[/tex]

What you did was increase the power rather than decreasing it.

In general, the derivative of secant raised to a power is ... [tex]\frac{d}{dx}(\sec^{n}x)=n\sec^{n}x\tan x[/tex]
i got the + sec^2x but how did you get the other + sec^2x
 

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