Question

Determine dy/dx for x*sin(y)=3

Answer 1

The answer that I got for this was dy/dx = -cos y ; but, I'm not really sure how to do this.

Answer 2

take the derivative of both sides, thinking that \(y=f(x)\) so you need both the product and the chain rule for the left
the derivative of the right is 0

Answer 3

\[x\sin(y)=3\]
\[\sin(y)+x\cos(y)y'=0\]

Answer 4

solve the equation for \(y'\) via
\[x\cos(y)y'=-\sin(y)\]
\[y'=-\frac{\sin(y)}{x\cos(y)}\]

Answer 5

like being asked for the derivative of
\[x\sin(f(x))\] need both product and chain rule

Answer 6

@satellite73-- What if I had something like:2x^2+x*y+3y^2=6 ?

Answer 7

then you need the product and chain rule again
like being asked for the derivative of
\[2x^2+xf(x)+3f^2(x)\]

Answer 8

Hmm. So, wherever I have a y in the equation, I can replace it with f(x)...

Answer 9

the derivative of \(2x^2\) is \(4x\)
the derivative of \(xf(x)\) is \(f(x)+xf'(x)\) by the product rule and the derivative of
\[3f^2(x)\] is \(6f(x)f'(x)\) again by chain rule

Answer 10

yeah that is my explanation not really what you want to do
you are thinking that \(y\) is a function of \(x\) even though you have not written it explicitly as one
that is why it is called "implicit" differentiation

Answer 11

\[2x^2+xy+3y^2=6 \]
\[4x+y+xy'+6yy'=0\] is a lot easier to write, than it is to replace \(y\) by \(f(x)\) and \(y'\) by \(f'(x)\)

Answer 12

That's what I had before with the y'

Answer 13

good

Answer 14

Would the answer be: -5x/6y? :O

Answer 15

oh no requires more algebra than that

Answer 16

Aw, I got excited for a minute.

Answer 17

\[4x+y+xy'+6yy'=0\]
\[xy'+6yy'=-4x-y\]
\[(x+6y)y'=-4x-y\]
\[y'=\frac{-4x-y}{x+6y}\]

Answer 18

it is just algebra though not calculus at that step

Answer 19

Ohh, so the derivative of whatever variable I'm taking stays to the left?

Answer 20

you have to solve for \(y'\) using algebra, no matter where it is

Answer 21

\[x^2+y^2=xy\]
\[2x+2yy'=y+xy'\] you still have to solve for \(y'\)