My Math Forum Chain rule to find dy/dt

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 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 = 3x²-10x 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 (3x²-10x)( -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
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Quote:
 Originally Posted by v8archie 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.
Determine the value of x how?

 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
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Quote:
 Originally Posted by v8archie You gave an expression for $x$ in terms of $t$ in your first post.
If I substitute t=-1 for x I get ( -1/2)
Then to find dy/dt do I simply miltiply (3x²-10x)( -1/2)?

February 1st, 2016, 09:38 AM   #8
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Quote:
 Originally Posted by atari So here is how far I got: dy/dx = 3x²-10x dx/dt = (t²+3)¹⁄²
You mean x= (t²+3)¹⁄², not dx/dt.
Quote:
 = t(t2+3)⁻¹⁄²
This is dx/dt

Quote:
 at t=-1 => -1/2 What do I do next, to find dy/dt, am I to multiply (3x²-10x)( -1/2)?Am I on the right track?

 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^2-10x dx/dt = t(t^2+3)^-1/2 dy/dt = (3x^2-10x)(t/sqrtt^2+3) substituting t=-1 (3x^2-10x)(-1/2)

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