Question

Find the equation of the line tangent to sin(xy)+y^4= 1 at (0,1).

Answer 1

know implicit differentiation ?

Answer 2

yes, have to solve for y prime. Im a little confused on what to do with the sin(xy), product rule or chain rule?

Answer 3

chain rule first, because its the function of function of x.
then u need product rule for xy,
try it

Answer 4

for the chain rule, would it be cos(xy) * y ' ?

Answer 5

should i give one next step ?

Answer 6

d/dx (sin(xy)) = cos (xy) d/dx(xy)
this is chain rule, got it ?

Answer 7

d/dx(xy) should be simple product rule.....

Answer 8

y cos (xy) ?

Answer 9

did u take y as constant and so (xy)' is just y ? NO, y is the function of x, so u need product rule like this :
(xy)' = x'y + xy'
now what is x' = ?

Answer 10

got that^ ?

Answer 11

lol, so sorry im just drawing blanks right now for some reason!

Answer 12

so should i give you solution so that u can look up or you want to try on your own first ?

Answer 13

hmmmm solution so I can see how its done if you dont mind.

Answer 14

sin(xy)+y^4= 1
diff. w.r.t x
(sin(xy)+y^4)'= 0
(sin(xy))'+(y^4)'= 0
cos(xy) (xy)' + 4y^3 y' =0
cos (xy) (x'y+y'x) +4y^3 y' =0
now x'=d/dx(x)=1
cos (xy) (y+y'x) +4y^3 y' =0
put point (0,1), that is put x=0,y=1
cos 0 = 1
(1+0) + 4 y' = 0
y' = -1/4 = slope of tangent at point (0,1)
clear till here ? ask if there is any doubt in any step.

Answer 15

cos(xy)(y+y'x)+4y^3y'=0 after that step I'm a little confused. how do you start to solve for y'.

Answer 16

the slope of tangent at a point (x1,y1) is given by putting x=x1, y= y1 in
y'= d/dx(f(x))
so i put y=1, x=0 in cos(xy)(y+y'x)+4y^3y'=0

Answer 17

okay, sorry it took so long but i see it now lol.

Answer 18

so now u have slope of line, there is a point on that line (0,1)
can u find the equation of line ?

Answer 19

one question first, could you factor out y' to help somewhat simplify?

Answer 20

like in cos(xy)(y+y'x)+4y^3*y'

Answer 21

oh u just want y' from cos(xy)(y+y'x)+4y^3*y' =0 ,right ?
y cos (xy) + y'(xcos xy) + y' (4y^3) = 0
can u find y' from here ?

Answer 22

xcosxy +4y^3 / ycos(xy) ?

Answer 23

y cos (xy) + y'(xcos xy) + y' (4y^3) = 0
y'(xcos xy) + y' (4y^3) = - y cos (xy)
y' = - y cos (xy) / (x cos xy +4y^3)

Answer 24

if u now put x=0, y=1, u still get y'=slope = -1/4

Answer 25

so would the tangent line be y-1=(-1/4)(x-0)

Answer 26

absolutely!
just simplify it ...

Answer 27

okay so y=-1/4x +1 ?

Answer 28

that is in slope intercept form,correct.
is it mentioned in which form the equation is needed ?

Answer 29

no but our professor used it in that form, because we have to approximate after solving for y.

Answer 30

then keep it in that form.
so u got all steps ?

Answer 31

yes lol. thank you so much, and sorry it took so long!

Answer 32

no problem .
welcome ^_^
have a nice day :)

Answer 33

you too! :)