Differentiation of a function wrt different variables

Question

Suppose a function "f (xy)", then does it equate to "f ' (xy) . x" when partial differentiating wrt y, and f '(xy).y when partial differentiating wrt x?

Cause I don't get it why differentiating f with x and y would both yield f ', any help?

Answer 1

Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.

Suppose that $$ f : \mathbb{R} \to \mathbb{R}$$ is a real-valued function of a single real variable, and let $$ u : \mathbb{R}^2 \to \mathbb{R} $$ be a real-valued function of two real variables defined by the formula $$ u = u(x,y) = xy.$$ Then the function $g = f\circ u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule: $$ g_x = \frac{\mathrm{d}(f\circ u)}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\mathrm{d}f}{\mathrm{d}u} \cdot y,$$ where $\frac{\mathrm{d}f}{\mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $\frac{\mathrm{d}f}{\mathrm{d}u}(u)$.

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Perhaps an example would help: suppose that $f(u) = \cos(u)$, and take $u = xy$. Then $$ \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{df}}{\mathrm{d}u} \cdot \frac{\mathrm{d}u}{\mathrm{d}x}. \tag{1}$$ But $$ \frac{\mathrm{d}f}{\mathrm{d}u} = \frac{\mathrm{d}}{\mathrm{d}u} \cos(u) = -\sin(u). $$ Then, as $u = xy$, it follows that $\frac{\mathrm{d}f}{\mathrm{d}u} = -\sin(xy)$. Also note that $\frac{\mathrm{d}u}{\mathrm{d}x} = y$. Substituting these into (1), we conclude that $$\frac{\mathrm{d}f}{\mathrm{d}x} = -\sin(xy) \cdot y. $$