Question

Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)

Answer 1

i got [sin(y) cos(x) + e^y sin(x)] / [e^y cos(x) - sin(x) cos (y)] but the textbook answer is [e^y sin(x) + y cos(xy)] / [e^y cos(x) - x cos(xy)] so am i close or am i there and if those answers are the same how are they equal show me

Answer 2

\[y'e^y \cos x -e^y \sin x=0+\cos(xy)(x)y'+\cos(xy)y\]so,\[y'=\frac{e^ysinx+\cos(xy)y}{e^ycosx-xcos(xy)}\]

Answer 3

not sure where you are getting siny*cosx terms...they are incorrect

Answer 4

why is d/dx [sin(xy)] equal to cos(xy)(x)y' + cos(xy)y ???? and i know the derivative of cos x, i had it as negative

Answer 5

i thought it was like sin(x)*sin(y)....... so that's how i tried to get the derivative oops but i still don't get what it is then

Answer 6

using the product rule I will differentiate
d/dx [sin(xy)] = d/dx [sin(xy)]*d/dx[xy]

Answer 7

we get:
cos (xy)*y

Answer 8

no ok i know how

Answer 9

but sin isn't its own fluttering thing

Answer 10

oh is it

Answer 11

oh wait flutter i'm reallll rusty on my trig i never learned it properly professor never got around to it so i had to cram myself for the departmental exam

Answer 12

so yeah thanks anyway

Answer 13

idk

Answer 14

this is the first term. but we have to do it again with respect to y and apply to chain rule so, the second term is:\[d/dx[\sin(xy)]=\cos(xy)*d/dy(xy)*dy/dx\]

Answer 15

looks messy but, you get:\[xcos(xy)*y'\]

Answer 16

now you sum them (this is just the usual product rule).

Answer 17

so really we are applying the product rule AND the chain rule to get this

Answer 18

wait WHAT the flutter d/dx (xy) is x*d/dx(y)(* dy/dx)+y*d/dx(x) so it's x(1) dy/dx + y(1) soooo the answer would be xcos(xy) dy/dx + ycos(xy)

Answer 19

as an easier example of this process consider differentiating xy implicitly.
d/dx[xy]=y+(d/dy)(dy/dx)(xy)
=y+xy'

Answer 20

why

Answer 21

maybe it's easier if we write
xy=xf(x)
where y= f(x)
now differentiate it:
d/dx[x*f(x)]=f(x)+xf'(x)
or,
=y+xy'

Answer 22

this is just the product rule, right? when we implicitly differentiate we assume that y is a function of x. y=f(x)

Answer 23

yeah yeah but flutter I DON"T EVEN KNOW THE flutterING IMPLICIT DIFFERENTIATION RULES MY TEXTBOOK SUCKS AND MY PROFESSOR SUCKS AND THERE ARE NO flutterING RULES IN THE TEXTBOOK FOR SOME REASON IT JUST SHOWS EXAMPLES WITHOUT EVEN EXPLAINING so ALL I KNOW is that you stick y' to multipy the derivative of any term with a y in it

Answer 24

that's fluttering it i don't know why or how or the details or fluttering anything flutter

Answer 25

when we have x*f(x) the overall change must take into account the rate of change of x and the rate of change of f(x). the product rule says that the overall rate of change of x*f(x) is the rate of change of x (i.e., =1) time f(x) PLUS x*(the rate of change of f(x)=f'(x))

Answer 26

and wtf so the product rule for sin(xy) how does that work too

Answer 27

product rule is d/dx[uv] = u d/dx[v] + v d/dx (u). wtf is u and what is v first of all.

Answer 28

remember we work from the outside to the inside. if we has sin(8x), working outside in, we get cos(8x)*d/dx(8x)=8cosx.

Answer 29

in sin(xy) how does that translate

Answer 30

but xy is x*y

Answer 31

ugh

Answer 32

yeah, the only difference between sin(8x) and sin(xy) is y is a function of x and 8 is not

Answer 33

so we have to account for this in the overall rate of change.
sin(xy)=cos(xy)*(d/dx)(xy)

Answer 34

now...the crucial thing to realize here is both x and y are functions of x in the above

Answer 35

so we have cos(xy) times d/dx(xy). we use the product rule on this second term.
d/dx(xy)=y+xy'
So overall we get
cos(xy)[y+xy']=ycos(xy)+xy'cos(xy

Answer 36

got to go, hope this helps!