Second Derivative of Trigonometric Functions

[tad.confused]

Homework Statement
Find the second derivative (y") of y=xtanx.


The attempt at a solution
I got the first derivative (y')
y=xtanx
y'=x(secx)+tanx

I started the second derivative and got stuck
y"=xsec^2x+tanx

 

[tiny-tim]

welcome to pf!

hi tad.confused! welcome to pf!

Find the second derivative (y") of y=xtanx.

I got the first derivative (y')
y=xtanx
y'=x(secx)+tanx

no, (tanx)' = sec²x

(if you're not convinced about that, apply the quotient rule to tanx = sinx/cosx)

and (secx)' = secx tanx

 

[JonF]

your y' is a bit off, it should be:

y'=x*sec²x+tanx

 

[blumist]

You got the first derivative completely wrong. Derivative of Tan x is not sec x, its sec^2 x. This way your problem goes as follows.

y = x tan x
y' = tanx + x sec^2 x

and

y" = sec²x + x ( 2 sec²x tan x ) + sec² x

[ Derivative sec²x is given by first reducing the SQUARE FORM to a simple linear form by the Power Rule of Derivatives and then consequently taking the derivative of the reduced function.

 

[tad.confused]

You got the first derivative completely wrong. Derivative of Tan x is not sec x, its sec^2 x. This way your problem goes as follows.

y = x tan x
y' = tanx + x sec^2 x

and

y" = sec²x + x ( 2 sec²x tan x ) + sec² x

[ Derivative sec²x is given by first reducing the SQUARE FORM to a simple linear form by the Power Rule of Derivatives and then consequently taking the derivative of the reduced function.

Oh, I had it right on the paper, but I typed the first derivative wrong. oops.

 

[tad.confused]

ok, so i get what was done, but the answer that is in the back of the book (and I understand that they are wrong sometimes) says that it is (2cosx+2xsinx)/((cos^3)x)

 

[tad.confused]

also, if I did the math right, I went from

(sec^2x+(2sec^2xtanx)+sec^2x)
and then dividing it all by sec^2x and canceled all of the sec^2x's and got 2+2xtanx

This is *kind of* close to what the back of my book says which is (2+2xtanx)/cos^2x=(2cosx+2xsinx)/((cos^3)x)

 

[tiny-tim]

hi tad.confused!

(try using the X² icon just above the Reply box )

and then dividing it all by sec^2x and canceled all of the sec^2x's and got 2+2xtanx

This is *kind of* close to what the back of my book says which is (2+2xtanx)/cos^2x=(2cosx+2xsinx)/((cos^3)x)

that is the same … just use secx = 1/cosx

(and blumist, please don't give out full answers … on this forum, we help people to get the answers themselves … see the two previous posts )

 

[tad.confused]

hi tad.confused!

(try using the X² icon just above the Reply box )


that is the same … just use secx = 1/cosx

(and blumist, please don't give out full answers … on this forum, we help people to get the answers themselves … see the two previous posts )

Oh! wow didn't even think about that! thanks!
Oh, so secx=1/cosx and its the same for sec²x=1/cos²x?
also, thank you for the tip about the subscript, so much easier

 

[tiny-tim]

Oh, so secx=1/cosx and its the same for sec²x=1/cos²x?

Yup!

(And now I think you now need to learn all the trigonometric identities in the PF Library )

 

[tad.confused]

Yup!

(And now I think you now need to learn all the trigonometric identities in the PF Library )

I get it now! Thank you sooooo much!

 

[blumist]

Sorry... Didnt really think about it!

You got it Tim.