Multivariable Chain Rule

Question

Use the chain rule to fine $\frac{\partial z}{\partial x}$ where $z=we^{4y}$, $w=2x^{0.5}$ and $y=\ln x$.

I found that $\frac {\partial z}{\partial x} = 31.5x^{2.5}$

Can someone check to see if this is correct?

Answer 1

We have that $z=z(w,y)$,$w=w(x)$, and $y=y(x)$ so $$\frac {\partial z}{\partial x}=\frac {\partial z}{\partial w}\frac {dw}{dx}+\frac {\partial z}{\partial y}\frac {dy}{dx}$$ then find all of the derivatives $$\frac {\partial z}{\partial w}=e^{4y}$$ $$\frac {dw}{dx}= \frac {1} {\sqrt x}$$ $$\frac {\partial z}{\partial y}=4we^{4y}$$ $$\frac {dy}{dx}= \frac 1x$$ so you get $$\frac {\partial z}{\partial x}= \frac {e^{4y}}{\sqrt x}+ \frac{4we^{4y}}{x}=\frac {e^{4 \ln x}}{\sqrt x}+ \frac{8 \sqrt xe^{4 \ln x}}{x}=x^{3.5}+8x^{3.5}= 9x^{3.5}$$

Answer 2

$z(x)=2\sqrt{x}\mathrm{e}^{4\ln x}=2\sqrt{x}x^4=2x^{9/2}$, so $$\frac{\mathrm{d}z}{\mathrm{d}x}(x)=9x^{7/2}.$$