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calyne
Group Title
Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)
 2 years ago
 2 years ago
calyne Group Title
Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)
 2 years ago
 2 years ago

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calyne Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

eseidl Group TitleBest 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,\[y'=\frac{e^ysinx+\cos(xy)y}{e^ycosxxcos(xy)}\]
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
not sure where you are getting siny*cosx terms...they are incorrect
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

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

eseidl Group TitleBest ResponseYou've already chosen the best response.0
we get: cos (xy)*y
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
no ok i know how
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
but sin isn't its own fluttering thing
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
so yeah thanks anyway
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
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:\[d/dx[\sin(xy)]=\cos(xy)*d/dy(xy)*dy/dx\]
 2 years ago

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

eseidl Group TitleBest ResponseYou've already chosen the best response.0
now you sum them (this is just the usual product rule).
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
so really we are applying the product rule AND the chain rule to get this
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
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)
 2 years ago

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

eseidl Group TitleBest ResponseYou've already chosen the best response.0
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'
 2 years ago

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

calyne Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
that's fluttering it i don't know why or how or the details or fluttering anything flutter
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
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))
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
and wtf so the product rule for sin(xy) how does that work too
 2 years ago

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

eseidl Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
in sin(xy) how does that translate
 2 years ago

calyne Group TitleBest ResponseYou've already chosen the best response.0
but xy is x*y
 2 years ago

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

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

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

eseidl Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

eseidl Group TitleBest ResponseYou've already chosen the best response.0
got to go, hope this helps!
 2 years ago
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