First, you must know these formula d(e^f(x) = f'(x)e^x dx, e^a+b=e^a.e^b, and d(sinx) = cosxdx, secx = 1/ cosx
(secx)dy/dx=e^(y+sinx), implies <span>dy/dx=cosx .e^(y+sinx), and then
</span>dy=cosx .e^(y+sinx).dx, integdy=integ(cosx .e^(y+sinx).dx, equivalent of
integdy=integ(cosx .e^y.e^sinx)dx, integdy=e^y.integ.(cosx e^sinx)dx, but we know that d(e^sinx) =cosx e^sinx dx,
so integ.d(e^sinx) =integ.cosx e^sinx dx,
and e^sinx + C=integ.cosx e^sinxdx
finally, integdy=e^y.integ.(cosx e^sinx)dx=e^2. (e^sinx) +C
the answer is
y = e^2. (e^sinx) +C, you can check this answer to calculate dy/dx
Answer: 21 dollars
Explain: 25 is too much and the next number down was good
This can be factored as (x-3)^2, so they are equal
Answer:
H
Step-by-step explanation:
