Question

find the dierivative using implicid differentiation: (e^y)sinx= x + xy

Answer 1

\[e^{y}\sin(x)=x+xy ?\]

Answer 2

that it yes!

Answer 3

ok cool do you know the derivatives of e^y and sin(x)?

Answer 4

e^y would be e^y, and sin(x) is cos(x)

Answer 5

sweet, and where are you getting stuck?

Answer 6

*** e^y would be e^y, and sin(x) is cos(x) ***
you mean

d/dx e^y = e^y dy/dx

d/dx sin(x) = cos(x) dx/dx = cos(x)

Answer 7

I don't know how to do this implicitly, like right now i have: (e^y)cosx=1+y+xy' and i don't know what to do now to solve in terms of y'

Answer 8

first do the left side

\[ \frac{d}{dx} (e^y \sin(x)) )\]


treat both y and x as variables.... that means you must use the product rule
the product rule is

d ( u v) = u dv + v du
be sure to use the chain rule d/dx e^y = e^y dy/dx (often written as e^y y' )

Answer 9

yah I'm lost

Answer 10

so would this side be : (e^y)cosx + e^yy' sinx ?

Answer 11

and the other 1+y+xy' ?

Answer 12

yes, that looks good.
now the right side
d(x + xy)= dx/dx + x dy/dx + y dx/dx
= 1 + x y' + y

yes

Answer 13

now solve for y'

Answer 14

okay, so now i have y' on both sides! what do i do?

Answer 15

(e^y)cosx + e^yy' sinx = 1 + x y' + y
(e^y)cosx-1-y = x y' - sin(x) e^y y'
factor out y'
divide both sides by (x- sin(x)e^y)

Answer 16

woahh i finally understand, I was thinking the y' was an exponent with the e! awesome! thanks so much! So my answer is y'=(1+y-e^y cosx)/((e^y)sin(x)-x)