Answer to Question #136776 in Calculus for Promise Omiponle

Question

Let z= 4e^x lny ,x= ln(ucosv), and y=usinv.  Calculate 𝜕z/𝜕u and 𝜕z/𝜕v first using the chain rule, and then by writing z(u,v) = z(x(u,v),y(u,v)) and taking partial derivatives directly. Show all steps each way

Answer 1

$z=4e^x\ln y, x=\ln(u\cos v), y=u\sin v.$

Calculate $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}$ using the chain rule:

$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial u}= 4e^x\ln y\frac{\cos v}{u\cos v}+4e^x\frac{1}{y}\sin v= 4e^{\ln(u\cos v)}\ln(u\sin v)\frac{1}{u}+4e^{\ln(u\cos v)}\frac{1}{u\sin v}\sin v = 4u\cos v\ln(u\sin v)\frac{1}{u}+4u\cos v\frac{1}{u}= 4\cos v\ln(u\sin v)+4\cos v$

$\frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}$

$\frac{\partial z}{\partial v}= 4e^x\ln y\frac{-u\sin v}{u\cos v}+4e^x\frac{1}{y}u\cos v= -4e^{\ln(u\cos v)}\ln(u\sin v)\frac{\sin v}{\cos v} +4e^{\ln(u\cos v)}\frac{1}{u\sin v}u\cos v= -4u\cos v\ln(u\sin v)\frac{\sin v}{\cos v} +4u\cos v\frac{1}{u\sin v}u\cos v=-4u\ln(u\sin v)\sin v+4u\frac{\cos^2 v}{\sin v}$

Write $z(u,v) = z(x(u,v),y(u,v))$ and take partial derivatives directly:

$z=4e^{\ln(u\cos v)}\ln(u\sin v)=4u\cos v\ln(u\sin v)$

$\frac{\partial z}{\partial u}= 4\cos v\ln(u\sin v)+4u\cos v\frac{\sin v}{u\sin v}= 4\cos v\ln(u\sin v)+4\cos v$

$\frac{\partial z}{\partial v}=-4u\sin v\ln(u\sin v)+4u\cos v\frac{u\cos v}{u\sin v}=-4u\sin v\ln(u\sin v)+4u\frac{\cos^2 v}{\sin v}$