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

- anonymous

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

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- schrodinger

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- 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

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

- anonymous

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

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## More answers

- 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

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

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

- anonymous

we get:
cos (xy)*y

- anonymous

no ok i know how

- anonymous

but sin isn't its own fluttering thing

- anonymous

oh is it

- 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

so yeah thanks anyway

- anonymous

idk

- 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

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

- anonymous

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

- anonymous

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

- 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

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

- anonymous

why

- 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

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

- 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

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

- 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

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

- 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

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

in sin(xy) how does that translate

- anonymous

but xy is x*y

- anonymous

ugh

- anonymous

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

- anonymous

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

- anonymous

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

- 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

got to go, hope this helps!

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