# First Derivative

1. Nov 5, 2008

### DollarBill

1. The problem statement, all variables and given/known data
Find the first derivative for

x sin(y) = y cos(x)

3. The attempt at a solution
x sin(y) = y cos(x)

Product rule on each side:

x cos(y) + sin(y) = -y sin(x) + cos(x)

I'm not sure what to do after that or if if I was even doing it right to begin with.

2. Nov 5, 2008

### HallsofIvy

What are you differentiating with respect to- x or y? It must be one or the other. If you are differentiating with respect to x, then the derivative of sin(y) is NOT cos(y), it is, by the chain rule, cos(y)(dy/dx) and the derivative of y is NOT 1, it is dy/dx. If you are differentiating with respect to y, then the derivative of x is NOT 1, it is dx/dy, and the derivative of cos(x) is NOT -sin(x), it is -sin(x)(dx/dy).

Finally, what does the problem really say? I am sure that it does not say "Find the first derivative for x sin(y) = y cos(x)", that doesn't make sense. I suspect it said "Find the derivatvive of y if x sin(y)= y cos(x)" though I can't be sure.

3. Nov 5, 2008

### DollarBill

This is all it says:

"Find the first derivative for
x sin(y) = y cos(x)"

It does have answer choices, but they don't have any equal signs in them. They don't say "dy/dx =" or "dx/dy =". Here are the answer choices:

A) Sin(x) - x sin(y) / y cos(x) - cos(y)

B) cos(x) - sin(y) / sin(x) - cos(y)

C) y sin(x) + sin(y) / cos(x) - x cos(y)

D) sin(x) + sin(y) / cos(x) + cos(y)

E) None of the above

4. Nov 5, 2008

### gabbagabbahey

I'm sure that they want you to find $dy \over{dx}$ by implicit differentiation.

5. Nov 5, 2008

### DollarBill

Working it out again for dy/dx:

dx/dx Sin(y) + x cos(y)dy/dx = dy/dx cos(x) - y sin(x)dx/dx

dx/dx just goes to 1:
Sin(y) + x cos(y)dy/dx = dy/dx cos(x) - y sin(x)

I'm pretty much where I was before except with a dy/dx on each side that would just cancel each other out.

6. Nov 5, 2008

### gabbagabbahey

The $dy \over{dx}$'s don't "cancel each other out". There are two terms on each side of your equation, and only one term per side has $dy \over{dx}$ as a factor...Try isolating the terms with $dy \over{dx}$ on one side of the equation, and the rest of the terms on the other.

7. Nov 5, 2008

### DollarBill

Sin(y) + x cos(y)dy/dx + y sin(x) / cos(x)= dy/dx

How would I get rid of the 2nd dy/dx?

8. Nov 5, 2008

### DollarBill

Well seeing as there's no dy/dx in the answer, I tried to do the following:

dx/dx Sin(y) + x cos(y)dy/dx = dy/dx cos(x) - y sin(x)dx/dx

dx/dx Sin(y) + x cos(y)dy/dx

dy/dx = -sin(y)/x cos(y)

dy/dx cos(x) - y sin(x)dx/dx

dy/dx=y sin(x) / cos(x)

-sin(y)/x cos(y) = y sin(x) / cos(x)

Cross multiply to get:

y sin(x)*x cos(y) / cos(x)*-sin(y)

But it's still not an answer choice

9. Nov 5, 2008

### gabbagabbahey

Let's start again from here:

$dy \over{dx}$ is a variable, and it happens to be the variable you are trying to solve for in the above equation.

Start by moving the terms that have $dy \over{dx}$ in them to one side of the equation and move the terms that don't have $dy \over{dx}$ in them to the other side.

For example, if I had the equation $Ax+sin(y) \frac{dy}{dx}=By\frac{dy}{dx}-3cos(x)$, I would solve it as follows:

Step 1; isolate the terms with $dy \over{dx}$:

$$Ax+sin(y) \frac{dy}{dx}=By\frac{dy}{dx}-3cos(x) \Rightarrow sin(y) \frac{dy}{dx}-By\frac{dy}{dx}=-Ax-3cos(x)$$

Step 2; factor out a $dy \over{dx}$:

$$sin(y) \frac{dy}{dx}-By\frac{dy}{dx}=-Ax-3cos(x) \Rightarrow \frac{dy}{dx}(sin(y)-By)=-Ax-3cos(x)$$

Step 3: divide by $(sin(y)-By)$ and hence solve for $dy \over{dx}$:

$$\frac{dy}{dx}=\frac{-Ax-3cos(x)}{(sin(y)-By)}$$

and so my answer would be $$\frac{-Ax-3cos(x)}{(sin(y)-By)}$$

Apply this method to your problem.

Last edited: Nov 5, 2008
10. Nov 5, 2008

### DollarBill

y sin(x) + sin(y) / cos(x)-x cos(y)

Wow, that was really easy. I was way over thinking that one :yuck:

Thanks

11. Nov 5, 2008

Yep