Chain Rule with Multivariable

Question

Using the chain rule, calculate $\dfrac{\mathrm{d}z}{\mathrm{d}x}$ of $z = we^{4y}$, where $w = 2\sqrt{x}$ and $y = \ln x$, and express it as a function of $x$ only.

I have tried to use $$\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{\mathrm{d}z}{\mathrm{d}w}\cdot\frac{\mathrm{d}w}{\mathrm{d}y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}$$ but I am having difficulty denoting $\dfrac{\mathrm{d}w}{\mathrm{d}y}$.

Answer 1

You can first express z in terms of x, making all of the substitutions in the beginning.

$z = 2\sqrt x e^{4\ln x}\\ z = 2x^{\frac 92}\\ \frac {dz}{dx} = 9x^\frac 72$

or you could use a combination of product rules and chain rules.

$\frac {d}{dx} z = $$(\frac {d}{dx} w) e^{4y} + w\frac {d}{dx}e^{4y}\\ \frac {dw}{dx}e^{4y}+ w(\frac {d}{dy} e^{4y})(\frac {dy}{dx})\\ \frac {1}{\sqrt x}e^{4y} + 8\sqrt xe^{4y}{\frac 1x}\\ \frac{9}{\sqrt x} e^{4y}\\ \frac{9}{\sqrt x} x^4\\ 9x^\frac 72$

Answer 2

It is: $$\frac{dz}{dx}=\frac{dz}{dw}\cdot \frac{dw}{dx}+\frac{dz}{dy}\cdot \frac{dy}{dx}=\\ 1\cdot e^{4y}\cdot \frac{1}{\sqrt{x}}+w\cdot 4e^{4y}\cdot \frac{1}{x}.$$