Question

using implicit differentiation find y prime if y=csc(xy)

Answer 1

i can do it until i get to the point where i have product rule for (xy) but once i get derivative of that (y-xy') i don't know how to get y' to other side

Answer 2

can you show me what you have so far?

Answer 3

when you use implicit differentiation you are still going to have y and x. you just want to make dy/dx be on a side by itself. when you have something like \[y ^{2}=x \]
you take the derivative of each side. this would be 2y dy/dx = 1 now divide by 2y and you get dy/dx = 1/2y

Answer 4

\[y'=-\csc(xy)\cot(xy)(y+xy')\]
\[y'=-y[\csc(xy)\cot(xy)]-xy'[\csc(xy)\cot(xy)]\]
\[y'+xy'\csc(xy)\cot(xy)=-ycsc(xy)\cot(xy)\]
\[y'(1+xcsc(xy)\cot(xy)=-ycsc(xy)\cot(xy)\]
\[y'=\frac{ -ycsc(xy)\cot(xy) }{1+xcsc(xy)\cot(xy)}\]

Answer 5

that what i get as my answer but the answer to the problem is -ycsc(xy)cot(xy) so basically my numerator

Answer 6

can you see any errors in my work @jeremyggg

Answer 7

just one moment i am working it out right now!

Answer 8

thank you so much!

Answer 9

i am getting the same as you

Answer 10

okay, I guess i'll just go with what i have then and hope i get partial marks. thanks!

Answer 11

@satellite73 can you please look over my workings as well?

Answer 12

sure