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Section 13.6 : Chain Rule
1. Given the following information use the Chain Rule to determine \(\displaystyle \frac{{dz}}{{dt}}\) .
\[z = \cos \left( {y\,{x^2}} \right)\,\hspace{0.5in}x = {t^4} - 2t,\,\,\,\,y = 1 - {t^6}\] Show SolutionOkay, we can just use the “formula” from the notes to determine this derivative. Here is the work for this problem.
\[\begin{align*}\frac{{dz}}{{dt}} & = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}}\\ & = \left[ { - 2xy\sin \left( {y{x^2}} \right)} \right]\left[ {4{t^3} - 2} \right] + \left[ { - {x^2}\sin \left( {y{x^2}} \right)} \right]\left[ { - 6{t^5}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 2\left( {{t^4} - 2t} \right)\left( {1 - {t^6}} \right)\left( {4{t^3} - 2} \right)\sin \left( {\left( {1 - {t^6}} \right){{\left( {{t^4} - 2t} \right)}^2}} \right) + 6{t^5}{{\left( {{t^4} - 2t} \right)}^2}\sin \left( {\left( {1 - {t^6}} \right){{\left( {{t^4} - 2t} \right)}^2}} \right)}}\end{align*}\]In the second step we added brackets just to make it clear which term came from which derivative in the “formula”.
Also, we plugged in for \(x\) and \(y\) in the third step just to get an equation in \(t\). For some of these, due to the mess of the final formula, it might have been easier to just leave the \(x\)’s and \(y\)’s alone and acknowledge their definition in terms of \(t\) to keep the answer a little “nicer”. You should probably ask your instructor for his/her preference in this regard.