DLS
  • DLS
siny=xsin(a+y) dy/dx=?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
I have a small doubt here..
DLS
  • DLS
is xsin(a+y) to be solved with product rule?
lgbasallote
  • lgbasallote
dy/dx means you take the derivative of x... since sin(a+y) does not include any x in it... no, you don't use product rule. you treat sin(a+y) as a constant coefficient

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

DLS
  • DLS
:o?
hartnn
  • hartnn
ummm.... u need to use product rule...
DLS
  • DLS
I did..
hartnn
  • hartnn
because y is the function of x
DLS
  • DLS
What I want to ask is..if I had ysin(a+y) then I wuldve used chain rule?
lgbasallote
  • lgbasallote
you only use product rule when two facots include x for example xsin x <-- x and sinx both have x so you use product rule xsiny <--- siny doesn't have x so treat it as coefficient remember that @hartnn ?
hartnn
  • hartnn
no, product again, because both are functions of x
hartnn
  • hartnn
sin y doesn't have x, right bu since y is the function of x....
hartnn
  • hartnn
d/dx sin y = cos y dy/dx
DLS
  • DLS
ive done that only
DLS
  • DLS
but even WFA treats it as 0
hartnn
  • hartnn
WFA ?
DLS
  • DLS
wolfaram :P ALPHA
DLS
  • DLS
spelling/*
hartnn
  • hartnn
no...x sin(a+y) needs product rule.
DLS
  • DLS
:/
hartnn
  • hartnn
x cos(a+y) dy/dx + sin(a+y)
hartnn
  • hartnn
cos y dy/dx =x cos(a+y) dy/dx + sin(a+y) isolate dy/dx
hartnn
  • hartnn
your WFA http://www.wolframalpha.com/input/?i=siny%3Dxsin%28a%2By%29%2C++dy%2Fdx%3D%3F
hartnn
  • hartnn
it doesn't treat it as 0.
DLS
  • DLS
oh nvm :o
DLS
  • DLS
but still not clear why product rule..its used in product of 2 functions so x and sin(a+y) x is not a function,y is
hartnn
  • hartnn
x is a function of x x^1
DLS
  • DLS
doesnt make any sense :S
DLS
  • DLS
every variable is afunction of itself then
hartnn
  • hartnn
you can say that. f(x) = x^n with n=1.
DLS
  • DLS
why didnt we use chain rule for sin(a+y)
hartnn
  • hartnn
we actually did.
DLS
  • DLS
:o
DLS
  • DLS
sin is one function and y other right?
hartnn
  • hartnn
cos (a+y) d/dx(a+y) =cos (a+y) (0+dy/dx)
hartnn
  • hartnn
yes.
DLS
  • DLS
so I have to treat x&y both as function not variables but other variable like abcd as constants(0) ?
DLS
  • DLS
when differentiating wrt x
anonymous
  • anonymous
\[\frac{d}{dx}xsin(a+y) \]\[=sin(a+y) \frac{d}{dx}(x) + x\frac{d}{dx}sin(a+y)\]\[=sin(a+y) + x[cos(a+y)\frac{d}{dx}(a+y)]\]\[=sin(a+y) + xcos(a+y)\frac{dy}{dx}\]
hartnn
  • hartnn
here since y is the function of x, we treat it as different function. if a is function of x, then a is also function..... if a is independent of x, then a is constant.
DLS
  • DLS
clear!
hartnn
  • hartnn
:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.