My Math Forum Chain rule to find dy/dt

 Calculus Calculus Math Forum

 February 1st, 2016, 05:18 PM #11 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 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 $\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=(3x^2-10x)\cdot\dfrac{-1}{2}=-\dfrac12(3(2)^2-10(2))=4$. Thanks from atari Last edited by greg1313; February 2nd, 2016 at 06:52 AM.
 February 2nd, 2016, 04:06 AM #12 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2215 At $t=-1, \ \dfrac{dx}{dt}=-\dfrac12$, not $-\dfrac14$. Also, $\dfrac{dy}{dx} = 3x^2 - 10x$, not $3x^2 - 5x$. Thanks from greg1313 and atari
 February 2nd, 2016, 06:03 AM #13 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 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: 20,966 Thanks: 2215 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: 7,963 Thanks: 1148 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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X=t² find dx/dt

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