Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculus Derivative Question

  1. Mar 28, 2005 #1

    Anyone out there able to help me out? I'm trying to find y'' of the equation y=xtanx. I found y' to equal 1(sec²x) but I don't know what to do after that. I know the final answer should be 2cosx + 2xsinx/ cos³x after being simplified and stuff, but I am clueless as to how to get there. Thanks in advance to anyone who can help me!
  2. jcsd
  3. Mar 28, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Nope,it can't be.Are u sure u want to compute

    [tex] \frac{d^{2}(x\tan x)}{dx^{2}} [/tex]...?

    Then better apply the Leibniz rule with care.

  4. Mar 28, 2005 #3
    You mean 2cosx + 2xsinx/ cos³x isn't the right answer or the y' I came up with to help me get the y'' isn't right?
  5. Mar 28, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Do you how to get to this result

    [tex] \frac{d(x\tan x)}{dx}=\tan x +x\sec^{2}x [/tex]...?

    If u do,i don't see any reason not to compute the 2-nd derivative correctly.

  6. Mar 28, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    Oh,your final aswer for the 2-nd derivative is in terms of "sin" & "cos",u'd better conver the 1-st derivative to a form containing "sin" a "cos" b4 the differentiation.

  7. Mar 28, 2005 #6
    Your first derivative is wrong. You need to use the product rule. Do you remember it?
  8. Mar 28, 2005 #7
    Data is right use the product rule to find first derivative and in finding the second derivative also.

    [tex] \frac{d(x\tan x)}{dx}=\tan x +x\sec^{2}x [/tex]

    [tex] \frac{d^2(x\tan x)}{dx^2}= \frac{(2)(\cos{x} +x\sin{x})}{\cos^3{x}} [/tex]
    Last edited: Mar 28, 2005
  9. Mar 28, 2005 #8
    indeed. Or for the second derivative you can just take it without converting to sines and cosines and get [itex]2\sec^2{x} + 2x\sec^2{x}\tan{x}[/itex], an equivalent answer.
  10. Mar 29, 2005 #9
    Where did you get the 2's from? Why isn't it just [itex]sec^2{x} + sec^2{x}\tan{x}[/itex]
  11. Mar 29, 2005 #10
    [tex] \frac{d}{dx} (\tan{x} + x\sec^2{x}) = \sec^2{x} + \frac{d}{dx} x\sec^2{x}[/tex]

    [tex] = \sec^2{x} + \left(\sec^2{x}\frac{d}{dx} x + x \frac{d}{dx}\sec^2{x}\right)[/tex] <------- Product Rule

    [tex] = \sec^2{x} + \left(\sec^2{x} + x\left(2(\sec{x}\tan{x})\sec{x}\right)\right)[/tex] <----- Chain Rule

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

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

    Remember, when you're taking the derivative of a product, you need to use the product rule:

    [tex] \frac{d}{dx} f(x)g(x) = f^\prime(x)g(x) + g^\prime(x)f(x) \neq f^\prime(x)g^\prime(x)[/tex]
    Last edited: Mar 29, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook