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I am not understanding what I'm doing wrong with the following question.

Suppose $w = \dfrac{x}{y} + \dfrac{y}{z}, x = e^t, y = 2 + \sin(t), z = 2 + \cos(4t)$. Find $\dfrac{dw}{dt}$ as a function of $x,y,z$ and $t$ and do no write $x,y,$ and $z$ in terms of $t$ nor rewrite $e^t$ as $x$.

My answer is $\dfrac{et}{y} - \dfrac{x\cos(t)}{y^2} + \dfrac{4y\sin(4t)}{z^2}$ which is the result of calculating $\dfrac{dw}{dt} = \dfrac{dw}{dx}\dfrac{dx}{dt} + \dfrac{dw}{dy}\dfrac{dy}{dt} + \dfrac{dw}{dz}\dfrac{dz}{dt}$.

Any help would be much appreciated. Thanks in advance.

letsmakemuffinstogether
  • 1,834

2 Answers2

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The answer is (the term you had missed out is highlighted in red) $$\begin{align}\dfrac{dw}{dt} &= \dfrac{dw}{dx}\dfrac{dx}{dt} + \dfrac{dw}{dy}\dfrac{dy}{dt} + \dfrac{dw}{dz}\dfrac{dz}{dt}\\&=\left(\frac{1}{y}\right)e^t+\left(-\frac{x}{y^2}+\color{red}{\frac{1}{z}}\right)\cos t+\left(-\frac{y}{z^2}\right)(-4\sin(4t))\\&=\frac{e^t}{y}-\frac{x\cos t}{y^2}+\frac{z\cos t+4y\sin (4t)}{z^2}\end{align}$$

Alijah Ahmed
  • 11,609
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Very close. You left out one term when doing

$$\frac{\partial w}{\partial y}\frac{dy}{dt}= \left[ -\frac{x}{y^2} +\frac{1}{z}\right] \cos(t) $$ So you need to add $\frac{\cos t}{z}$ to your answer.

By the way, you can edit my answer to see how to write the "partial" symbol $\partial$.

Mark Fischler
  • 41,743