Calculus halp

this is what i got as well. The answer this sheet gives me is( -tan^2e^x)/(sine^y) hopefully it's wrong because i can't get it
 
tan(e^x)-x = cos(e^y)

ok, so you would get.... e^xsec^2(e^x)-1 = e^y (dy/dx)(-sin(e^y))

and then solve for dy/dx:

(e^(x-y)sec^2(e^x)-1)/(-sin(e^y)) = dy/dx

anyone else get that? Im not sure if that's right
 
same, i was in a calc class that I never did homework or studied in and still got and A. That being said most the people in my class got Cs
 
Just kidding. It is differentiating shit from like calc 1. I completely forget how to do that. But you probably have to use trig differentiations and chain rule.
 
Yeah ^ that seems about right, but you forgot the implicit part with dy/dx and stuff, but knowing what variable you're doing with this with respect to would help. I'd guess x, you the ys need to be dy/dx right? Haven't done one of those a while.
Calc related tangent (it's like a pun or something)
I'm deep into my power series homework problems, I did the ratio test for
∞∑ _[(-1)^n](x-1)^n n=1 (3^2n)(ln(n))
Wow that was hard to type. But did the ratio test for that and I forgot to write down the (x-1)^n and (x-1)^(n+1) terms in the absolute value ratio thing, finished, then realized that I forgot those. fuuuuuuuuuuuuuuuuuu
 
(e^x)sec^2(e^x)-1=(e^y)dy/dx(-sin(e^y))

then solve for dy/dx

i think thats right, i could quite possibly be wrong though
 
yup that's what I got, but somehow it's supposed to be

( -tan^2e^x)/(sine^y).... I don't remember all my trig identities so I can't get any farther than solving for dy/dx with what you got.
 
You must log in or register to reply here.
Back
Top