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How would I do the following problem


$\frac{dy}{dt}$ when $t=1$

if $y=x^2+3x-10$ and $x=\frac{4-t}{3+t}$

I know that when $t=1$ $x=3/4$

So I took the derivative of $x^2+3x-10$ and got $2x+3$ and plug in $3/4$ for $x$ and got $9/2$ then I took the derivative of $\frac{4-t}{3+t}$ and got

$\frac{(3+t)(-1)-(1)(4-t)}{(3+t)^2}$ plug in $t=1$ and got $\frac{-7}{16}$ so would I multiply ${(9/2)}{(-7/16)}$ to get the correct answer.

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Write $y$ as a function of $t$ just substituting $x$ and then use the chain rule to derivate. – Sigur Feb 14 '13 at 19:23
It looks and is very messy solution but it's correct. Next step for you is to look for some solved examples in any book and try to write solutions in similar style (more clearly). – zaarcis Feb 14 '13 at 19:26
Looks good to me! – hardmath Feb 14 '13 at 19:26
I hope I did it correctly. – Fernando Martinez Feb 14 '13 at 19:29
up vote 1 down vote accepted

Yes, it's correct.

But you could have written $y$ as a function of $t$ by substituting the expression given for $x$ into that equation, and then used the chain-rule.

E.g., Evaluate $\frac{dy}{dt}$ when $t=1$

if $y=x^2+3x-10$ and $x=\frac{4-t}{3+t}$, then $$y = \left(\frac{4-t}{3+t}\right)^2 + 3\left(\frac{4-t}{3+t}\right) - 10$$

Then computed $\frac{dy}{dt}$ and evaluated at the value at $t=1$.

But you got the job done! (You'll find, if you haven't already, that there are often many ways to approach the same problem!)

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Thanks the way you wrote the problem with substitution is more efficient I think. – Fernando Martinez Feb 14 '13 at 19:54

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