Results 1 to 4 of 4

Math Help - Question x^x

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    49

    Question x^x

    If <br />
\frac{{df}}{{dx}}=x^x <br />
for x>0, and  <br />
g(x)=f(x^2) then <br />
\frac{{dg}}{{dx}} =<br />
?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chaddy View Post
    If <br />
\frac{{df}}{{dx}}=x^x <br />
for x>0, and  <br />
g(x)=f(x^2) then <br />
\frac{{dg}}{{dx}} =<br />
?
    by the chain rule: \frac {dg}{dx} = f'(x^2) \cdot 2x

    now continue
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    49
    Okay, so <br />
\frac {dg}{dx} = f'(x^2) \cdot 2x<br />
which means that <br />
\frac {dg}{dx} = x^{2x^2} \cdot 2x<br />
    Thus <br />
\frac {dg}{dx} = 2x^{2x^2+1}<br />
    Correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by chaddy View Post
    Okay, so <br />
\frac {dg}{dx} = f'(x^2) \cdot 2x<br />
which means that <br />
\frac {dg}{dx} = x^{2x^2} \cdot 2x<br />
    Thus <br />
\frac {dg}{dx} = 2x^{2x^2+1}<br />
    Correct?
    yup
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum