Question

Please help again. What is dy/dx of x=sin(xy) in steps. I am missing something! Thanks

Answer 1

This looks like a job for implicit differentiation and the chain rule.

Answer 2

Yep and I am making an error somewhere in the chain rule.

Answer 3

Well, it also includes some product rule.

d/dx [sin(xy)] = ?

Answer 4

This is the chain rule step. d/dx [sin(xy)] = ? What have you got?

Answer 5

To respond you just reply to your own message.

Answer 6

I am guessing you went wrong with the chain rule when you got to the product xy.

d/dx (xy) = ?

Answer 7

Okay, you messaged me that d/dx (xy) is xy' + 1y.

[I'm trying not to get slapped for not Socraticing you again.]

So

d/dx [sin(xy)] = cos(xy)(xy' + y)

that combines the chain rule and the product rule

From the beginning:

x=sin(xy)
d/dx x = d/dx [sin(xy)]

1 = cos(xy)(xy' + y)

now you want to solve for y' because y' is dy/dx.

Answer 8

\(\displaystyle \begin{align} {1 \over {\cos(xy)}} &= xy^\prime + y \cr {1 \over {\cos(xy)}} -y &= xy^\prime \cr \left( {1 \over {\cos(xy)}} -y \right)({1 \over x}) &= y^\prime \cr \end{align}\)

Answer 9

check that so far.

You were given \(\large x=\sin(xy) \)

Answer 10

See if you can make it agree with your textbook answer if you substitute for x.

Answer 11

You message me that your textbook answere is
\(\displaystyle y^\prime ={1 \over {y\cos(xy)}}\)

Answer 12

\(\displaystyle \begin{align} \left( {1 \over {\cos(xy)}} -y \right)({1 \over x}) &= y^\prime \cr \left( {1 \over {\cos(xy)}} -y \right) \left( {1 \over {\sin(xy)}} \right) &= y^\prime \cr \left( {1 \over {\cos(xy)\sin(xy)}} \right) - \left( {y \over {\sin(xy)}} \right) &= y^\prime \cr \end{align}\)

Answer 13

Aside from expressing the left with a common denominator, I am out of ideas here.

Answer 14

As I am too!