Find $\frac{\partial f}{\partial x}$ when:
$f = 3xy^2z^3$, $y = 3x^2 + 2$, $z = \sqrt{x-1}$
I would need to replace $y$ and $z$ in $f$ before calculating the partial of $f$, correct?
(The alternative would be find the partial of $y$ and $z$, square and cube them respectively and multiply all that by 3, which I think is incorrect, but I'm not sure)
On
We are free to either replace $y=y(x)$ and $z=z(x)$ or keep them in the calculation. If we decide to keep them in, which might also be convenient, we have to be aware that both $y$ and $z$ are functions of $x$ and have to treat them accordingly.
We consider \begin{align*} f(x,y,z)&=3xy^2z^3\\ y&=y(x)=3x^2+2\\ z&=z(x)=\sqrt{x-1} \end{align*} Differentiating $y=y(x)$ and $z=z(x)$ we obtain \begin{align*} y^\prime(x)=6x\qquad\qquad z^\prime(x)=\frac{1}{2\sqrt{x-1}} \end{align*}
One way to do the partial differentiation is applying the product rule as following \begin{align*} \color{blue}{\frac{\partial}{\partial x}f(x,y,z)}&=((3xy^2)z^3)^{\prime}\\ &=(3xy^2)^\prime z^3+3xy^2(z^3)^\prime\\ &=(3y^2+3x\cdot 2yy^\prime)z^3+3xy^2\cdot 3z^2z^\prime\\ &\,\,\color{blue}{=3y^2z^3+6xyy^\prime z^3+9xy^2z^2z^\prime}\tag{1}\\ &=3(3x^2+2)^2(x-1)^{\frac{3}{2}}+6x(3x^2+2)6x(x-1)^{\frac{3}{2}}\\ &\qquad+9x(3x^2+2)^2\frac{1}{2}(x-1)^{\frac{1}{2}}\\ &\,\,\color{blue}{=\frac{3}{2}(x-1)^{\frac{1}{2}}(3x^2+2)(39x^3-30x^2+10x-4)} \end{align*}
Similarly we obtain
\begin{align*} \frac{\partial}{\partial x}f(x,y,z)&=(3x(y^2z^3))^{\prime}\\ &=(3x)^\prime y^2z^3+3x(y^2z^3)^\prime\\ &=3y^2z^3+3x(2yy^\prime z^3+3y^2z^2z^\prime)\\ &=3y^2z^3+6xyy^\prime z^3+9xy^2z^2z^\prime\\ \end{align*} giving the same result as in (1).
On
It is probably wanting you to use the Chain Rule for partial derivatives with intermediates. This rule is, if $u = f(x, y)$, $x = g(s, t)$, and $y = h(s, t)$, then
$$ \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s}$$
So, in your case, this would mean:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial x}$$
However, the $\frac{\partial f}{\partial x}$ is actually the total differential, $\frac{df}{dx}$. $\frac{\partial f}{\partial x}$ refers to different things on each side of the equal sign. On the left, it is the grand-total partial differential of $f$ with respect to $x$ (which, in this case, is the same as the total differential as well). On the right, it is the partial differential of $f$ with respect to $x$ only as far as $x$ is concerned by itself.
Yes, but it wouldn't be partial. $y$ and $z$ are both functions of $x$, so it doesn't really make sense to treat them as constants. So you would have $$f(x)=3x(3x^2+2)(x-1)^{3/2}=3(3x^3+2)(x-1)^{3/2}$$ which you can easily differentiate using the Product Rule.