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'