anonymous
  • anonymous
Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
\[y'e^y \cos x -e^y \sin x=0+\cos(xy)(x)y'+\cos(xy)y\]so,\[y'=\frac{e^ysinx+\cos(xy)y}{e^ycosx-xcos(xy)}\]
anonymous
  • anonymous
not sure where you are getting siny*cosx terms...they are incorrect

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anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
using the product rule I will differentiate d/dx [sin(xy)] = d/dx [sin(xy)]*d/dx[xy]
anonymous
  • anonymous
we get: cos (xy)*y
anonymous
  • anonymous
no ok i know how
anonymous
  • anonymous
but sin isn't its own fluttering thing
anonymous
  • anonymous
oh is it
anonymous
  • anonymous
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
anonymous
  • anonymous
so yeah thanks anyway
anonymous
  • anonymous
idk
anonymous
  • anonymous
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:\[d/dx[\sin(xy)]=\cos(xy)*d/dy(xy)*dy/dx\]
anonymous
  • anonymous
looks messy but, you get:\[xcos(xy)*y'\]
anonymous
  • anonymous
now you sum them (this is just the usual product rule).
anonymous
  • anonymous
so really we are applying the product rule AND the chain rule to get this
anonymous
  • anonymous
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)
anonymous
  • anonymous
as an easier example of this process consider differentiating xy implicitly. d/dx[xy]=y+(d/dy)(dy/dx)(xy) =y+xy'
anonymous
  • anonymous
why
anonymous
  • anonymous
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'
anonymous
  • anonymous
this is just the product rule, right? when we implicitly differentiate we assume that y is a function of x. y=f(x)
anonymous
  • anonymous
yeah yeah but flutter I DON"T EVEN KNOW THE FUCKING IMPLICIT DIFFERENTIATION RULES MY TEXTBOOK SUCKS AND MY PROFESSOR SUCKS AND THERE ARE NO FUCKING 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
  • anonymous
that's fluttering it i don't know why or how or the details or fucking anything fuck
anonymous
  • anonymous
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))
anonymous
  • anonymous
and wtf so the product rule for sin(xy) how does that work too
anonymous
  • anonymous
product rule is d/dx[uv] = u d/dx[v] + v d/dx (u). wtf is u and what is v first of all.
anonymous
  • anonymous
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.
anonymous
  • anonymous
in sin(xy) how does that translate
anonymous
  • anonymous
but xy is x*y
anonymous
  • anonymous
ugh
anonymous
  • anonymous
yeah, the only difference between sin(8x) and sin(xy) is y is a function of x and 8 is not
anonymous
  • anonymous
so we have to account for this in the overall rate of change. sin(xy)=cos(xy)*(d/dx)(xy)
anonymous
  • anonymous
now...the crucial thing to realize here is both x and y are functions of x in the above
anonymous
  • anonymous
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
anonymous
  • anonymous
got to go, hope this helps!

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