# Can normal chain rule be used for total derivative?

Chain rule states that $$\frac{df}{dx} \frac{dx}{dt} = \frac{df}{dt}$$.

Suppose that $f$ is function $f(x,y)$. In this case, would normal chain rule still work?

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The multivariable chain rule goes like this:

$$\frac{df}{dt} = \frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$$ If you can isolate for $\dfrac{dy}{dx}$, then you can always just do implicit differentiation.

Let's do an example: $$f=f(x,y) = x^2 - y$$ Where $$x(t) = t, \; \; y(t) = t$$ $$\frac{df}{dt} \frac{df}{dx} \cdot \frac{dx}{dt} = 2x -\frac{dy}{dx}\frac{dx}{dt}= 2t - 1$$ Also $$\frac{df}{dt} = \frac{\partial f}{\partial x}\cdot \frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt} = 2x - 1 = 2t - 1$$ Now let's do the same example, except this time: $$f=f(x,y)= x^2 - y\\ x(t)=t\ln(t), \;\; y(t) = t\sin(t) \\ \Rightarrow \frac{df}{dt} = 2t\ln(t)(\ln(t) + 1) -(\sin(t) +t\cos(t))$$ Would implicit differentiation work here? Yes. But it requires more steps.
Implicit differentiation would be the inefficient route as, $$e^{x/t} = t \Rightarrow y= e^{x/t}\sin(t) \Rightarrow \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} = \\ \left(\frac{1}{t}\frac{dx}{dt}-\frac{x}{t^2}\right)e^{\frac{x}{t}}\sin(t)+e^{\frac{x}{t}}\cos(t) = \\ e^{\frac{x}{t}}\left(\left(\frac{1}{t}\frac{dx}{dt}-\frac{x}{t^2}\right)\sin(t)+\cos(t)\right) =\\ t\left(\frac{\ln(t)+1}{t} - \frac{t\ln(t)}{t^2}\right)\sin(t) + t\cos(t) = \\ \left(\ln(t) +1 - \ln(t)\right)\sin(t) + t\cos(t) = \\ \sin(t)+t \cos(t)$$ But it still works out.

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No, actually, your calculation of $\frac{df}{dx} \cdot \frac{dx}{dt}$ is wrong - thinking of $df/dx$ is $(2x-dy/dx)$ and as $dy/dx = 1$ and $dx/dt = 1$, we get the same answer $2t-1$... –  Lanz Feb 4 '13 at 6:04
but $x=y=t$. Look, they must be 1. –  Lanz Feb 4 '13 at 6:09