Help on using the chain rule

Hi everyone :).

I'm looking at the below question and I don't understand its answer. I normally use Leibniz notation (chain rule) but I don't get how it is done here.

http://i.imgur.com/mxIxSfq.jpg

If someone could talk me through it, that would be great.

Thanks and sorry for the poor condition of the link.

f(x , y) = 1 / sqrt(x^2 + y^2)

We want to find df/dy, so yes you would just use the chain rule which gives you answer D.

Basically just rewrite the sqrt as an exponent:

f(x , y) = (x^2 + y^2)^(-1/2)

Now apply the chain rule by first taking the derivative of the "outer function", which is the -1/2 exponent part:

(-1/2)*(x^2 + y^2)^(-3/2)

And then multiplying the derivative (with respect to y) of the inner function which is x^2 + y^2.

(-1/2)*(x^2 + y^2)^(-3/2) * (2y)

Simplifying, the -1/2 and the 2 cancel out to give:

(-y)*(x^2 + y^2)^(-3/2) = (-y) / (x^2 + y^2)^(3/2) --> Answer D


So the problem is either you got confused on some step or you don't fully understand how the chain rule works. In detail you can think about it like this:

We have some function f(x , y) and we want to find df/dy, but the form of f(x , y) is a bit complicated. What we realize is that f(x , y) is really a "function within a function" in the sense that it can be written as:

f(x , y) = f( h(x , y) )

In the example you gave,
h(x , y) = (x^2 + y^2) and f(h) = h^(-1/2)

The chain rule says that

df/dy = df/dh * dh/dy

So first we have to find df/dh and then multiply by dh/dy, because the "dh" differentials cancel to give us df/dy.

df/dh = (-1/2)*h^(-3/2) -- simple power rule
dh/dy = 2y -- power rule here also

df/dh * dh/dy = (-1/2)*h^(-3/2) * 2y = (-y)*h^(-3/2) = df/dy

Now finally you plug back in for h(x, y) to get a function of x and y.