January 31st, 2016, 06:38 PM  #1 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0  Chain rule to find dy/dt
Hello again, I am stuck at the question below and can't make out what exactly I need to do. I have been asked to use the Leibnitz notation of the chain rule to find dy/dt at t = −1 for y = x³ − 5x² and x =√ (t² + 3) Am I to somehow combine y and x and then substitute t? Would anyone be kind enough to clarify this for me? Last edited by skipjack; February 2nd, 2016 at 04:02 AM. Reason: to insert missing parentheses 
January 31st, 2016, 07:25 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675 
dy/dt = (dy/dx)(dx/dt)

January 31st, 2016, 08:09 PM  #3 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 
So here is how far I got: dy/dx = 3x²10x dx/dt = (t²+3)¹⁄² = t(t2+3)⁻¹⁄² at t=1 => 1/2 What do I do next, to find dy/dt, am I to multiply (3x²10x)( 1/2)?Am I on the right track? 
January 31st, 2016, 09:02 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra 
You are on the right track. Your next step should be determine the value of $x$ when $t=1$ and put that into your expression.

January 31st, 2016, 10:00 PM  #5 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0  
January 31st, 2016, 10:10 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra 
You gave an expression for $x$ in terms of $t$ in your first post.

January 31st, 2016, 10:15 PM  #7 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0  
February 1st, 2016, 09:38 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  
February 1st, 2016, 01:35 PM  #9 
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675 
$\displaystyle x=\sqrt{t^2+3}$ At t=1, x=2. 
February 1st, 2016, 03:47 PM  #10 
Member Joined: Jan 2016 From: Canada Posts: 40 Thanks: 0 
I guess got this wrong then. This was a hard concept to grasp. The answer I submitted was this: dy/dx = 3x^210x dx/dt = t(t^2+3)^1/2 dy/dt = (3x^210x)(t/sqrtt^2+3) substituting t=1 (3x^210x)(1/2) 

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