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### Section 3.11 : Related Rates

1. In the following assume that $$x$$ and $$y$$ are both functions of $$t$$. Given $$x = - 2$$, $$y = 1$$ and $$x' = - 4$$ determine $$y'$$ for the following equation.

$6{y^2} + {x^2} = 2 - {x^3}{{\bf{e}}^{4 - 4y}}$

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Hint : This is just like the problems worked in the section notes. The only difference is that you’ve been given the equation and all the needed information and so you won’t have to worry about finding that.
Start Solution

The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to $$t$$.

$12y\,y' + 2x\,x' = - 3{x^2}x'{{\bf{e}}^{4 - 4y}} + 4{x^3}{{\bf{e}}^{4 - 4y}}y'$ Show Step 2

All we need to do now is plug in the given information and solve for $$y'$$.

$12y' + 16 = 48 - 32y'\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{y' = {\textstyle{8 \over {11}}}}}$