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Say I have a function $f(x,y,z)$. If I know $t= \sqrt{x+\sqrt{x^2+ y*z}}$ and I know the partials $\large\frac{\partial{f}}{\partial{x}}$,$\large\frac{\partial{f}}{\partial{y}}$,$\large\frac{\partial{f}}{\partial{z}}$, how could I apply the chain rule in order to obtain $\large\frac{\partial{f}}{\partial{t}}$?

I would have thought:

$$\frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{t}} +\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{t}}$$

however, I already feel like i'm on the wrong track. Can anyone give me a start on how to construct $\large\frac{\partial{f}}{\partial{t}}$ in terms of $\large\frac{\partial{f}}{\partial{x}}$,$\large\frac{\partial{f}}{\partial{y}}$ and $\large\frac{\partial{f}}{\partial{z}}$?

user1952500
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adrien k
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1 Answers1

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Your deduction is formally correct i.e. it is true that $$ \frac{\partial{f}}{\partial{t}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{t}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{t}} +\frac{\partial{f}}{\partial{z}}\frac{\partial{z}}{\partial{t}}\label{1}\tag{1} $$ However, you missed a basic step: with your definition of $t$, how can you deduce three functions $$ t\mapsto \big(x(t), y(t), z(t)\big)\; $$ i.e. define a function $\boldsymbol{p}:\mathbb{R}\to \mathbb{R}^3$ such that $f(t)\triangleq f(\,\boldsymbol{p}(t))=f\big(x(t), y(t), z(t)\big)$?
The relation you are considering, i. e. $$ t= \sqrt{x+\sqrt{x^2+ y\cdot z}}\label{2}\tag{2} $$ defines a 2-dimensional variety in $\mathbb{R}^3$, and in order to parametrize it two parameters are needed, therefore you cannot use \eqref{2} to define any such $\boldsymbol{p}=\boldsymbol{p}(t)$.

Daniele Tampieri
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  • Is this true?: $$\frac{\partial{x}}{\partial{t}} = \frac{1}{\frac{\partial{t}}{\partial{x}}}$$ – user1952500 Nov 07 '18 at 20:59
  • You can calculate tangents to that variety, but how did you define $$\frac{\partial f}{\partial{x}}(x(t),y(t),z(t))?$$ It is still the dependence on the gradient of $f$ on $t$ which is unresolved. – Daniele Tampieri Nov 07 '18 at 21:13
  • And you can also think of using the relations $$\frac{\partial x}{\partial t}=\frac{1}{\frac{\partial t}{\partial x}}\quad \frac{\partial y}{\partial t}=\frac{1}{\frac{\partial t}{\partial y}}=\frac{\partial z}{\partial t}$$ to define a map from $\mathbb{R}\to\mathbb{R}^3$ by integrating them, but the integration constants depend on (at least one) of the remaining variables. – Daniele Tampieri Nov 07 '18 at 21:23