Question

Use implicit differentiation to find dy/dx.

y cos (1/y) = 8x + 8y

Answer 1

ok, so ignore my first comment. you have formula for that , tell me what you have from school

Answer 2

Im not sure what you are asking, I would start with getting the derivative and adding dy/dx to each y term. Then factoring out the dy/dx

Answer 3

yes, do it. I check

Answer 4

I am not sure how to get the derivative of the inside of the cos(). thats the part that is throwing me off

Answer 5

dy/dx -sin(1/y) = 8 + 8 dy/dx

Answer 6

you have quotient rule. the leftside has 2 terms, break it down and take derivative each of part. about the cos ( ) . it's is the "2 layers" derivative. the outer is cos, take derivative of cos to get -sin and then * (time ) to inner function's derivative , it 's is (1/y) '. do you know what I mean?

Answer 7

now, I use y' to indicate to dy/dx. the left side, you have [y cos(1/y) ]'= y'cos(1/y) + y (cos (1/y))'

Answer 8

so I need to use the quotient rule on the inside of -sin?

Answer 9

Im sorry this has me backwards at the moment

Answer 10

the quotient rule for the whole thing. the cos ( ) part need "2 layers" rule.

Answer 11

so the first term is y. the derivative of that is simply dy/dx. my second term is cos(1/y) and the der. is [-sin(1/y) - cos(1/dy/dx)] / y^2 ?

Answer 12

hey. your original one doesn't have - between y and cos

Answer 13

check it please. it's another topic,

Answer 14

the derivative of cos is -sin correct?

Answer 15

ycos( ) is a term need using quotient rule. OMG you confuse me

Answer 16

sorry.. imagine how i feel

Answer 17

Refer to the attachment, a solution using Mathematica for the calculations.

Answer 18

Perfect. thank you robtobey. @nicedart take a look at that attachment. helpful

Answer 19

thank you. I have a few more of these to do this should get me going!

Answer 20

You'all welcome.