Results 1 to 3 of 3

Thread: Check my answers for Higher derivatives

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    197

    Exclamation Check my answers for Higher derivatives

    1) find y' if $\displaystyle y=2sin2x+5cos3x-4$
    Attempt:
    $\displaystyle (2)sin(2x)+(5)cos(3x-4) $
    $\displaystyle =4cos2x-15sin3x-4$
    2) find y" if $\displaystyle y=cotx $
    Attempt:
    y'=$\displaystyle -csc^2x$
    y"=$\displaystyle 2csc^2xcotx$
    3) find y" if $\displaystyle y=x.tanx $
    Attempt:
    y'=$\displaystyle (x)(sex^2x)+tanx $
    y"=$\displaystyle (x)(-2sec^2x)tanx+sec^4x $
    4)find y" if $\displaystyle 3sin^3(2x) $
    Attempt;
    y'=$\displaystyle 18(sin^2x)(cos2x)$
    y"=$\displaystyle -36 sin^3 2x+4cos^2 2x sin2x $
    5)find y" if $\displaystyle x^5+3x^3+6 $
    Attempt:
    y'=$\displaystyle 5x^4+9x^2$
    y"=$\displaystyle 20x^3+18x$

    Please check is my answer are correct,if not write the correct solution in steps...
    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288
    You should use this site to check your derivatives

    Wolfram Mathematica Online Integrator

    For no.4 $\displaystyle y'$ should be $\displaystyle 18(sin^22x)(cos2x)$
    and
    $\displaystyle y''=18(2cos^22xsin2x+-2sin^32x)$

    No.3
    $\displaystyle y'=(x)(sex^2x)+tanx$
    $\displaystyle y''=2xsec^3xsinx+2sec^2x$

    N.B
    $\displaystyle \frac{d}{dx}(xsec^2x)=2xsec^3xsinx+sec^2x$
    Last edited by I-Think; Nov 11th 2009 at 05:24 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    197
    Quote Originally Posted by I-Think View Post
    You should use this site to check your derivatives

    Wolfram Mathematica Online Integrator

    For no.4 $\displaystyle y'$ should be $\displaystyle 18(sin^22x)(cos2x)$
    and
    $\displaystyle y''=18(2cos^22xsin2x+-2sin^32x)$

    No.3
    $\displaystyle y'=(x)(sex^2x)+tanx$
    $\displaystyle y''=2xsec^3xsinx+2sec^2x$

    N.B
    $\displaystyle \frac{d}{dx}(xsec^2x)=2xsec^3xsinx+sec^2x$
    So are there any mistakes ,btw thnx for the site but i am differentiating not integrating
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives: check my answers please
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 10th 2010, 01:02 PM
  2. Higher Derivatives r''(q)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 7th 2010, 10:22 AM
  3. Higher Derivatives
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Mar 5th 2009, 02:25 AM
  4. Derivatives CHECK ANSWERS
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 28th 2009, 01:59 AM
  5. Derivatives - Please Check my answers
    Posted in the Calculus Forum
    Replies: 12
    Last Post: Oct 18th 2007, 08:02 PM

Search Tags


/mathhelpforum @mathhelpforum