I found this exercise on the book of multivariable calculus from which I'm studying:

"Find the partial derivative $\frac{\partial{z}}{\partial{x}}$ and the total derivative $\frac{\text{d}z}{\text{d}x}$ of $z(x,y)=e^{xy}$ where $y=\phi(x)$."

Now, this to me looks like a function of a single variable $f:\mathbb{R}\to\mathbb{R}$ and so in this case the partial derivative of $f$ with respect to $x$ and total derivative would be equivalent; in particular, I end up with something like:


In the solution, while the result for the total derivative is the same as mine, the partial derivative of $f$ with respect to $x$ is written as follows:


Why is this the case? Since the partial derivative of $f$ with respect to $x$ shows the incremental behaviour of the function as $x$ changes, shouldn't I account for the presence of $x$ in the functional representation of $y$ while computing the derivative with respect to $x$?

Sorry in advance for the super basic question :)


It is possible for $\phi(x)$ to be stationary with respect to $x$ wiggles while $f(x,\phi(x))$ is not stationary due to its remaining explicit dependence on $x$. When we write the partial derivative, we by-definition require that $y$ is constant, or equivalently that $\phi(x)$ is constant while we wiggle x, so it doesn't contribute to the variation (hence why it gets called "partial"). The "total" variation of $f$ makes no such restriction, and thus depends on the general wiggle of both x and $\phi(x)$.

I assume you have an intuition / understanding of the following equality, $$ \frac{df(x,y)}{dt} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} $$

Letting $x=t$ we have, \begin{align} \frac{df(x,y)}{dx} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x}\\ &= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x} \end{align}

Letting $y = \phi(x)$ we have, $$ \frac{df(x,y)}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \phi'(x) $$

Notice that, $$ \phi'(x)=0\ \implies\ \frac{df}{dx}=\frac{\partial f}{\partial x} $$

The partial derivative under question is "how $f$ varies while we vary $x$ and hold $y$ constant", $\frac{\partial f}{\partial x}$, which still makes sense even if $y=\phi(x)$.

In the case of $f(x,y) = e^{xy}$ we have, $$ \frac{\partial f}{\partial x} = ye^{xy} $$ and \begin{align} \frac{df(x,y)}{dx} &= ye^{xy} + xe^{xy}\phi'(x)\\ &= e^{xy}\big{(}y + x\phi'(x)\big{)} \end{align}

which is, as you have found and we would hope, equivalent to thinking of $f$ as a function of one variable $x$ to find by "product rule", $$ \frac{df(x)}{dx} = e^{x\phi(x)}\big{(}\phi(x) + x\phi'(x)\big{)} $$

  • $\begingroup$ I think I got it, thank a lot! $\endgroup$ – Prometheus Dec 6 '18 at 12:37
  • $\begingroup$ @Stefano No problem. Generally, you can "accept" an answer to your Stack Exchange question by clicking the checkmark by the voting buttons. $\endgroup$ – jnez71 Dec 6 '18 at 15:39

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