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In this Multivariate Calculus course, in the partial derivatives part, the lecturer solves differentiates this function:

$$ f(x, y, z) = \sin(x)e^{yz^2} $$

He lists the partial derivatives: $$ \frac{\partial f}{\partial x} = \cos(x)e^{yz^2};\\ \frac{\partial f}{\partial y} = \sin(x)e^{yz^2}z^2;\\ \frac{\partial f}{\partial z} = \sin(x)e^{yz^2}2yz; $$

I understand how these are calculated. Then (as part of teaching the "total derivative" concept), he introduces the fact that $x$, $y$, $z$ are themselves a function of another parameter $t$, such that: $$ x = t-1;\\ y = t^2;\\ z=1/t; $$ Aside of simply substituting each term in the original expression with $t$ and then differentiating w.r.t $t$, he goes on (here) to differentiate each of the partial derivatives (but now, w.r.t $t$):

$$ \dfrac{dx}{dt} = 1;\\ \dfrac{dy}{dt} = 2t;\\ \dfrac{dz}{dt} = -t^{-2}; $$

I don't understand the final part. For example, how did we get to $\dfrac{dx}{dt} = 1$? If we take the partial derivative $\frac{\partial f}{\partial x} = \cos(x)e^{yz^2}$, substitute $x$ with $t-1$ and differentiate w.r.t $t$, shouldn't we get $\frac{dx}{dt} = -\sin(t-1)e^{yz^2}$?

Ricky
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HeyJude
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    Since $x(t)=t-1$ we have $\frac{dx}{dt} = \frac{d}{dt}(t-1) = \frac{d}{dt}t- \frac{d}{dt}1 = 1 - 0 =1$ – jjagmath Oct 04 '23 at 10:35
  • Oh, so we don't actually differentiate the partial derivative. Just the $x, y, z$ terms themselves as a function. ok, I got that part wrong, thanks! – HeyJude Oct 04 '23 at 10:39

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