Answer to Question #339351 in Calculus for FEL

Question #339351

Consider the function g defined by:


g(x,y) = cos ( y√y ) + 1/log3(x - y)


Determine ∂2g/∂y∂x



1
Expert's answer
2022-05-10T23:11:10-0400

Function g(x,y)g(x,y) has the form: g(x,y)=cos(xπy)+1log3(xy)=cos(xπy)+ln(3)ln(xy)g(x,y)=\cos(x\pi\sqrt{y})+\frac{1}{log_3(x-y)}=\cos(x\pi\sqrt{y})+\frac{\ln(3)}{ln(x-y)}.

We receive:

gy=xπ2ysin(xπy)+ln(3)(ln(xy))21xy\frac{\partial g}{\partial y}=-\frac{x\pi}{2\sqrt{y}}\sin(x\pi\sqrt{y})+\frac{\ln(3)}{(\ln(x-y))^2}\cdot\frac{1}{x-y}\\. 2gyx=π2ysin(xπy)xπ22cos(xπy)(ln(3))(2+ln(xy))(ln(xy))31(xy)2\,\frac{\partial^2g}{\partial y\partial x}=-\frac{\pi}{2\sqrt{y}}\sin(x\pi\sqrt{y})-\frac{x\pi^2}{2}\cos(x\pi\sqrt{y})-\frac{(\ln(3))(2+\ln(x-y))}{(\ln(x-y))^3}\cdot\frac{1}{(x-y)^2}.

Answer: 2gyx=π2ysin(xπy)xπ22cos(xπy)(ln(3))(2+ln(xy))(ln(xy))31(xy)2\frac{\partial^2g}{\partial y\partial x}=-\frac{\pi}{2\sqrt{y}}\sin(x\pi\sqrt{y})-\frac{x\pi^2}{2}\cos(x\pi\sqrt{y})-\frac{(\ln(3))(2+\ln(x-y))}{(\ln(x-y))^3}\cdot\frac{1}{(x-y)^2}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment