February 1st, 2016, 05:18 PM  #11 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
At $t=1$, $x=\sqrt{(1)^2+3}=2$. $\dfrac{dx}{dt}=\dfrac{t}{\sqrt{t^2+3}}$. At $t=1$, $\dfrac{dx}{dt}=\dfrac12$. So we have $\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=(3x^210x)\cdot\dfrac{1}{2}=\dfrac12(3(2)^210(2))=4$. Last edited by greg1313; February 2nd, 2016 at 06:52 AM. 
February 2nd, 2016, 04:06 AM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,648 Thanks: 2085 
At $t=1, \ \dfrac{dx}{dt}=\dfrac12$, not $\dfrac14$. Also, $\dfrac{dy}{dx} = 3x^2  10x$, not $3x^2  5x$. 
February 2nd, 2016, 06:03 AM  #13 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
Thanks. I've edited my post.

February 2nd, 2016, 06:50 AM  #14 
Global Moderator Joined: Dec 2006 Posts: 20,648 Thanks: 2085 
Your final answer now has the wrong sign.

February 2nd, 2016, 06:53 AM  #15 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 
Ok. Edited.

February 2nd, 2016, 09:41 AM  #16 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 
So my answer was right , I just did not simplify


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