Question

if 1/x+1/y=3 and y(4)=4/11, find y'(4) by implicit differentiation.

for the answer i got it was 0, i need help

Answer 1

imply it then :)

Answer 2

\[Dx(1/x)+Dx(1/y)=Dx(3` )\]

Answer 3

\[-x'.1/x^2-y'.1/y^2=x'.0\]

Answer 4

d/dx(1/x)=-1/x^2 and d/dx(1/y)=y'/y^2

1/x^2-y'/y^2

Answer 5

very good so far :)

Answer 6

yeah that's what i got
then i apply the (4, 4/11)

i get -1/16 and -y'/(16/121)

Answer 7

-1/16-y'/(16/121)

Answer 8

\[y'=-\frac{y^2}{x^2}=-\left(\frac{y}{x}\right)^2\]

Answer 9

i got .47265625 in fractions its 121/256 is that right

Answer 10

\[y' = \frac{16/121}{16/1}=\frac{1}{121}\]right?

Answer 11

hold on let me do it again

Answer 12

and i dropped the negative on accident lol

Answer 13

ohh sorry i flipped the 121and 16 yeah 1/121 is right

Answer 14

-1/121

Answer 15

yeah -1/121 is right thanks for the help

our professor said it equals 0 on another similar problem in class but the y' formula helps