A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)
anonymous
 4 years ago
Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'e^y \cos x e^y \sin x=0+\cos(xy)(x)y'+\cos(xy)y\]so,\[y'=\frac{e^ysinx+\cos(xy)y}{e^ycosxxcos(xy)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not sure where you are getting siny*cosx terms...they are incorrect

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0using the product rule I will differentiate d/dx [sin(xy)] = d/dx [sin(xy)]*d/dx[xy]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but sin isn't its own fluttering thing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so yeah thanks anyway

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this 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:\[d/dx[\sin(xy)]=\cos(xy)*d/dy(xy)*dy/dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0looks messy but, you get:\[xcos(xy)*y'\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you sum them (this is just the usual product rule).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so really we are applying the product rule AND the chain rule to get this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as an easier example of this process consider differentiating xy implicitly. d/dx[xy]=y+(d/dy)(dy/dx)(xy) =y+xy'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe 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'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is just the product rule, right? when we implicitly differentiate we assume that y is a function of x. y=f(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's fluttering it i don't know why or how or the details or fluttering anything flutter

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when 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))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and wtf so the product rule for sin(xy) how does that work too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0product rule is d/dx[uv] = u d/dx[v] + v d/dx (u). wtf is u and what is v first of all.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0remember we work from the outside to the inside. if we has sin(8x), working outside in, we get cos(8x)*d/dx(8x)=8cosx.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in sin(xy) how does that translate

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, the only difference between sin(8x) and sin(xy) is y is a function of x and 8 is not

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so we have to account for this in the overall rate of change. sin(xy)=cos(xy)*(d/dx)(xy)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now...the crucial thing to realize here is both x and y are functions of x in the above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0got to go, hope this helps!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.