Question

Implicit differentiation question

Answer 1

\[y = \sin xy\]

Answer 2

I get y' = cos xy[(xy')+(1y)]
y' = x(cos xy)y' + (cos xy)y
y' - x(cos xy)y' = (cos xy)y

Answer 3

What is the derivative dy/dx of sin x?

Now suppose that you have the separate function xy as the argument of the sin function: y=sin xy. First, you need to find the derivative of y=sin xy with respect to xy; then you multiply that derivative by the derivative of the argument xy.

xy is a product. How would you find the derivative of xy with respect to x?

Answer 4

\[\large\rm \color{orangered}{\text{Welcome to OpenStudy! :)}}\]

Hmm your steps look good so far!

Answer 5

But I don't get how you get from
y' - x(cos xy)y' = (cos xy)y
to
\[y'x = \frac{ \cos (xy)y }{ 1 - x(cost xy) }\]

Answer 6

Factoring! :)

\(\large\rm y'-y'(stuff)=y'(1-stuff)\)

Pull the y' out of each term, ya?

Answer 7

"y' = cos xy[(xy')+(1y)]" looks good. I would write it as

(dy/dx) = cos xy (d/dx) (xy).

= cos xy [x (dy/dx) + y (1) ].

First, let's back up and review the original question. What were you asked to find, specifically?

In my previous comments I've assumed you wanted to find dy/dx. Is that the case, or not?

Answer 8

Yeah I need to find dy/dx

Answer 9

Then solve (for y') the expression zepdrix has given you. At this point you have an algebra problem, since you've already found the necessary derivatives.

Answer 10

Where are you in this process?

Answer 11

I don't understand what y' - y'(stuff) = y' (1-stuff) relates to

Answer 12

like wheres the 1- ?

Answer 13

All right. Let's find the 'stuff.'

\[y = \sin xy\]
Goal: find y', or (dy/dx).

Answer 14

Find the derivative of both sides with respect to x.

Answer 15

y' - y'*stuff is the same as y'*1 - y'*stuff

There is a coefficient of 1 on that first term.
So when you divide y' out of each term, a 1 is left over in that first slot, ya?

Answer 16

would you be good enough to define for scrivmj exactly where that "stuff" came from? If you don't, I will.

Answer 17

Looks like we need to take smaller steps here.

Answer 18

Find the derivative of both sides with respect to x.

Answer 19

Be sure to use the chain rule for that separate function xy.

Answer 20

y = sin xy
y' = cos xy[xy'+1y]
y' = x(cos xy)y' + (cos xy)y
y' - x(cos xy)y' = (cos xy)y
y'[1 - x(cos xy)] = (cos xy)y
y'= (cos xy)y/[1 - x(cos xy)]

Answer 21

yayy good job c:

Answer 22

haha thanks

Answer 23

What is your goal at this point?
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Answer 24

the book says that's the final answer. you just have to find dy/dx by implicit differation

Answer 25

Agreed. We have to apply several rules, such as that for the derivative of the sine function, the chain rule (due to that separate function xy). (dy/dx) will show up on both sides of your equation, so you have to go thru the algebra to solve the equation for (dy/dx).

Answer 26

You have handled this well.

I'm an advocate of showing all work, which you've done in your final appraoch.

Answer 27

Reminder: Be certain to include the specific instructions you are to follow: "Use implicit differentiation to find (dy/dx)."