# Is this derivative correct

1. Aug 18, 2008

### asi123

1. The problem statement, all variables and given/known data

Guys, is this right?
And if it is, from the the y' got in there?

2. Relevant equations

3. The attempt at a solution

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2. Aug 18, 2008

### rootX

Re: Derivative

You should use imageshack/other for images

3. Aug 18, 2008

Re: Derivative

The answer looks reasonable - perhaps multiply the two negatives.
I'm not sure what you mean by "And if it is, from the the y' got in there?"

4. Aug 18, 2008

### HallsofIvy

Staff Emeritus
Re: Derivative

Here, z and y are both functions of some other variable, perhaps x or t. If z= f(y) and y is itself a function of x, then, by the chain rule
$$\frac{dz}{dx}= \frac{dz}{dy}\frac{dy}{dx}$$

If, in particular, z= sin(1/y)= sin(y-1), and y is a function of x, then
$$\frac{dz}{dx}= cos(1/y)(-1/y^2)\frac{dy}{dz}$$
or, the ' notation,
z'= cos(1/y)(-1/y2)y'= (-cos(1/y)/y2)y'