Results 1 to 8 of 8

Math Help - Derivative of 2^(xtanx)?

  1. #1
    Banned
    Joined
    Oct 2009
    Posts
    56

    Derivative of 2^(xtanx)?

    I don't even know where to begin on this one. It's not (xtanx)2^(xtanx-1) is it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    648
    Hello, satx!

    You're expected to know this differentiation formula:

    . . \text{If }\,f(x) \;=\;b^{u},\,\text{ then: }\:f'(x) \;=\;b^u\,u' \,\ln b


    f(x) \;=\;2^{x\tan x}

    We have: . f'(x) \;=\;2^{x\tan x}\cdot(x\sec^2x + \tan x)\cdot \ln 2

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,780
    Thanks
    1519
    Quote Originally Posted by satx View Post
    I don't even know where to begin on this one. It's not (xtanx)2^(xtanx-1) is it?
    No, it isn't. The "power law" only applies when the base is a variable and the power a constant.

    You need to know, first, that the derivative of a^x is ln(a) a^x. Further, by the chain rule, the derivative of a^u(x) is ln(a) a^{u(x)}\frac{du}{dx}.

    In this case, of course, u(x)= x tan(x). Use the product rule to find du/dx.

    (Once again, Soroban beats me by 1 minute!!!)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member mathemagister's Avatar
    Joined
    Feb 2010
    Posts
    191
    Quote Originally Posted by satx View Post
    I don't even know where to begin on this one. It's not (xtanx)2^(xtanx-1) is it?
    Use:
    The Exponential Rule

    \boxed{\frac{d}{dx}b^x = b^x \cdot \ln{b}}

    And the chain rule in the exponential rule:

    \frac{d}{dx}b^{f(x)} = b^{f(x)} \cdot \ln{b} \cdot f'(x)

    In your problem, b = 2 and f(x) = x \tan(x)

    Now use the product rule to find f'(x):

    f'(x) = x\sec^2(x) + \tan(x) \cdot 1

    So the answer is 2^{x\tan(x)}\cdot \ln2 \cdot (x\sec^2(x) + \tan(x))

    Hope that helps

    Mathemagister

    PS Sorry for the post taking so long and overlapping others' posts (my internet disconnected)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    56
    Merci beaucoup a tous!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Some of the posters in this thread seem to think that calculus is an exercise in the memorisation of a fist full of "rules".

    Here you need the definition of logarithms and a bit of manipulation:

    a^b=e^{\ln(a)b}

    Then the rest follows from the systematic application of the product and chain rules and a table of basic derivatives.

    Brain storage space is to precious to waste on large list of rules, understanding acts as a compression algorithm, the more you understand the less you need to remember.

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member mathemagister's Avatar
    Joined
    Feb 2010
    Posts
    191
    Quote Originally Posted by CaptainBlack View Post
    Some of the posters in this thread seem to think that calculus is an exercise in the memorisation of a fist full of "rules".

    Here you need the definition of logarithms and a bit of manipulation:

    a^b=e^{\ln(a)b}

    Then the rest follows from the systematic application of the product and chain rules and a table of basic derivatives.

    Brain storage space is to precious to waste on large list of rules, understanding acts as a compression algorithm, the more you understand the less you need to remember.

    CB
    Yes, but it is much easier to teach that way. Even though one understands the underlying meaning, it is more comprehensible for the person in doubt if they are taught in terms of the rules, which are a sort of the language in which we explain how to do a problem which is then translated back by the other person (hopefully) in such a way that they understand the underlying meaning.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by mathemagister View Post
    Yes, but it is much easier to teach that way. Even though one understands the underlying meaning, it is more comprehensible for the person in doubt if they are taught in terms of the rules, which are a sort of the language in which we explain how to do a problem which is then translated back by the other person (hopefully) in such a way that they understand the underlying meaning.
    I'm sorry but I don't agree, for instance I know there is a quotient rule, but do I know what it is? No because it is actually easier to use the product rule once the chain rule is internalised. Rule that are only occasionally used are more trouble that they are worth.

    Also rule based rote learning gives the wrong idea to the student of what mathematics is. I don't care if the student manages to clear artificial hurdles what I do care about is that they learn to appreciate what maths is about (beauty and pattern) and can use what they do know in novel situations.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. Replies: 2
    Last Post: November 6th 2009, 02:51 PM
  4. |xtanx| <= |x|
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 8th 2009, 01:52 AM
  5. Fréchet derivative and Gâteaux derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2008, 04:40 PM

Search Tags


/mathhelpforum @mathhelpforum