using total derivative to approximate changes
Sophia Terry 
I have the function $ f(x,y,z) = xyz$ with $y = x^2 $, $z = \sqrt[3]{x} $
The initial values are $ (27, 729, 3) $
I want to compute the total derivative of f wrt x and use that to approximate the change in f due to an increase in x by 0.1 units.
I am confused because my answer is not equivalent to what I get when I simply make $f = x^{7/2} $ and compute $ f'(x)*0.1 $.
The two different methods should give the same answer, so I think I am making a mistake with my calculation of the total derivative.
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$\begingroup$Your point given $(27,729,3)$ does not satisfy the condition on $z$, so I will use $z=\sqrt[3]{x}$ instead of $\sqrt{x}$.
Firstly, consider $f(x,y,z)=xyz$, with $y=x^2$ and $z=\sqrt[3]{x}$. Then $$ df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz=yz\,dx+xz\,dy+xy\,dz. $$ We are given $dx=0.1$, but we need to calculate $dy$ and $dz$. Using the same approach, $$ \begin{align} dy&=\frac{dy}{dx}dx=2x\,dx,\\ dz&=\frac{dz}{dx}dx=\frac{1}{3\sqrt[3]{x^2}}\,dx. \end{align} $$ This gives $dy=5.4$ and $dz=1/270$. Substituting all of this information into $df$ gives $df=729$.
The second method, with the adjusted $z$ function gives $f(x)=x^{10/3}$. Its differential is $$ df=\frac{df}{dx}dx=\frac{10}{3}x^{7/3}\,dx. $$ Substituting $x=27$ and $dx=0.1$ gives $df=729$.
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