# Is this derivative correct

## Homework Statement

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

## The Attempt at a Solution

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## Homework Statement

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

## The Attempt at a Solution

You should use imageshack/other for images

Homework Helper

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?"

HallsofIvy
Homework Helper

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'