Question

If 1/x+1/y=4 and y(4)=4/15, find y′(4) by implicit differentiation

Answer 1

\[\frac{1}{x}+\frac{1}{y}=4\]?

Answer 2

if so get
\[-\frac{1}{x^2}-\frac{1}{y^2}y'=0\]

Answer 3

solve for y' via
\[-\frac{1}{y^2}y'=\frac{1}{x^2}\]
\[y'=-\frac{y^2}{x^2}\]

Answer 4

replace y by 4 and x by whatever it has to be to make
\[\frac{1}{x}+\frac{15}{4}=4\]

Answer 5

then x=4 too :/

Answer 6

not getting it, man.

Answer 7

y'(4) is the value of y' when x = 4

so y'(4) = -y^2/16

now, to find y when x = 4, well that's already been given: y(4) = 4/15

hence y'(4) = -(4/15)^2 / 16

= -16/16(15)^2

= -1/225

Answer 8

ahh, I see. Cheers!