 multivariable functions works when the partial derivatives are continous but hvacwk 2022-01-07 Answered
multivariable functions works when the partial derivatives are continous but if the function is just differentiable does the chain rule work ? I mean if $$\displaystyle{z}={f{{\left({x},{y}\right)}}}$$ is differentiable function and $$\displaystyle{x}={g{{\left({t}\right)}}},{y}={h}{\left({t}\right)}$$ are also differentiable functions can we write:
$$\displaystyle{\frac{{{\left.{d}{z}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{\partial{f}}}{{\partial{x}}}}\times{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}+{\frac{{\partial{f}}}{{\partial{y}}}}\times{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}$$

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Stella Calderon

The derivative, if it exists, of a function $$\displaystyle{f}:\mathbb{R}^{{n}}\rightarrow\mathbb{R}^{{m}}$$ at a point x is a linear transformation $$\displaystyle{f}'{\left({x}\right)}:\mathbb{R}^{{n}}\rightarrow\mathbb{R}^{{m}}$$ that satisifies
$$\displaystyle{f{{\left({x}+{h}\right)}}}-{f{{\left({x}\right)}}}={f}'{\left({x}\right)}{h}+{r}{\left({h}\right)}$$ where $$\displaystyle{r}={o}{\left({\left|{h}\right|}\right)}$$
Now, suppose, $$\displaystyle\phi:\mathbb{R}\rightarrow\mathbb{R}^{{2}}:{t}\mapsto{\left({g{{\left({t}\right)}}},{h}{\left({t}\right)}\right)}$$ and $$\displaystyle{f}:\mathbb{R}^{{2}}\rightarrow\mathbb{R}$$. If $$\displaystyle\phi$$ is differentiable at $$\displaystyle{t}={t}_{{0}}$$ and f is differentiable at $$\displaystyle{\left({g{{\left({t}_{{0}}\right)}}},{h}{\left({t}_{{0}}\right)}\right)}$$, then we can apply the chain rule to assert that $$\displaystyle{f}\circ\phi:\mathbb{R}\rightarrow\mathbb{R}$$ is differentiable at $$\displaystyle{t}={t}_{{0}}$$ and
$$\displaystyle{\left({f}\circ\phi\right)}'{\left({t}_{{0}}\right)}:\mathbb{R}\rightarrow\mathbb{R}$$ exists and is equal to $$\displaystyle{f}'{\left(\phi\right)}{\left({t}_{{0}}\right)}{)}\circ\phi'{\left({t}_{{0}}\right)}$$
So far, nothing to do with partials.
Of course, it turns out that if the derivative exists then it matches your formula.
And if the relevant partial derivatives are continuous, the derivative exists and the formula is also satisfied.
On the other hand, if one or more of the partials is not continuous at a given point, the derivative may or may not exist there, and you need to check directly to see if (1) is satisfied.