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I'd like to compute $\frac{\partial x}{\partial z}$ along $S$ at $(x,y,z)$ for $S: \frac{1}{x}+\arctan(y+2z)=1$.

My Approach: I can define $w(x,y,z)=\frac{1}{x}+\arctan(y+2z)$ and find the total differential and so on, i.e., $dw=w_x dx+w_y dy+w_z dz$ (we'd also need to use the fact that $y$ is held constant and $dw=0$).

How can I use the multivariable chain rule here? I'd like to find $\frac{\partial x}{\partial z}$ using the chain rule, but I'm a little bummed out here because I am only used to using the chain rule for solving equations where, say, $y$ depends on $a,b$ and $a, b$ depend on $t$ (e.g., $\frac{dy}{dt}=\frac{\partial y}{\partial a}\frac{da}{dt}+\frac{\partial y}{\partial b}\frac{db}{dt}$).

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1 Answer 1

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We can write \begin{align*} S:\frac{1}{x}+\arctan(y+2z)=1\tag{1} \end{align*} as function in $x=x(y,z)$.

We obtain from (1) \begin{align*} x(y,z)&=\frac{1}{1-\arctan(y+2z)}\\ \\ \color{blue}{\frac{\partial x}{\partial z}(z,y)} &=\frac{\partial}{\partial z}\left(\frac{1}{1-\arctan(y+2z)}\right)\\ &=\frac{\frac{\partial}{\partial z}\left(\arctan(y+2z)\right)}{\left(1-\arctan(y+2z)\right)^2}\tag{2}\\ &=\frac{2}{\left(1+(y+2z)^2\right)\left(1-\arctan(y+2z)\right)^2}\\ &\,\,\color{blue}{=\frac{2x^2}{1+(y+2z)^2}}\tag{3} \end{align*}

Comment:

  • In (2) we use $\left(\frac{1}{g(z)}\right)^{\prime}=-\frac{\left(g(z)\right)^{\prime}}{(g(z))^2}$.

  • In (3) we use $x=\frac{1}{1-\arctan(y+2z)}$.

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