siny=xsin(a+y)
dy/dx=?

- DLS

siny=xsin(a+y)
dy/dx=?

- jamiebookeater

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- DLS

I have a small doubt here..

- DLS

is xsin(a+y) to be solved with product rule?

- 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

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## More answers

- DLS

:o?

- hartnn

ummm.... u need to use product rule...

- DLS

I did..

- hartnn

because y is the function of x

- DLS

What I want to ask is..if I had ysin(a+y) then I wuldve used chain rule?

- 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

no, product again, because both are functions of x

- hartnn

sin y doesn't have x, right
bu since y is the function of x....

- hartnn

d/dx sin y = cos y dy/dx

- DLS

ive done that only

- DLS

but even WFA treats it as 0

- hartnn

WFA ?

- DLS

wolfaram :P ALPHA

- DLS

spelling/*

- hartnn

no...x sin(a+y) needs product rule.

- DLS

:/

- hartnn

x cos(a+y) dy/dx + sin(a+y)

- hartnn

cos y dy/dx =x cos(a+y) dy/dx + sin(a+y)
isolate dy/dx

- hartnn

your WFA
http://www.wolframalpha.com/input/?i=siny%3Dxsin%28a%2By%29%2C++dy%2Fdx%3D%3F

- hartnn

it doesn't treat it as 0.

- DLS

oh nvm :o

- 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

x is a function of x
x^1

- DLS

doesnt make any sense :S

- DLS

every variable is afunction of itself then

- hartnn

you can say that.
f(x) = x^n with n=1.

- DLS

why didnt we use chain rule for sin(a+y)

- hartnn

we actually did.

- DLS

:o

- DLS

sin is one function and y other right?

- hartnn

cos (a+y) d/dx(a+y)
=cos (a+y) (0+dy/dx)

- hartnn

yes.

- DLS

so I have to treat x&y both as function not variables but other variable like abcd as constants(0) ?

- DLS

when differentiating wrt x

- 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

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

clear!

- hartnn

:)

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