Why is the power rule for derivatives not valid here? I am stuck on an exercise where I have to figure out the derivative of $y = \frac{\sqrt{20-x^2}}{4}$. I realize that this equation can be rewritten as:
$1/4 * \sqrt{20-x^2}$, so when I factor out the constant 1/4 on beforehand, this leaves me with $\sqrt{20-x^2}$. This can be rewritten as a power $(20-x^2)^{0.5}$. 
My idea was to simply apply the power rule to this equation to find the derivative. This leads to:
$0.5(20-x^2)^{-0.5} = \frac{1}{2} * \frac{1}{\sqrt{20-x^2}} = \frac{1}{2\sqrt{20-x^2}}$
However my solution is not valid, and wolfram alpha does something complicated using the chain rule instead. Please help me understand why. 
 A: You can employ the power rule, but you forget that in doing so, you're also invoking the chain rule, with kernel $(20-x^2)$. That means you have to multiply your result with the derivative of this expression as well, which is $-2x$. That should give you the answer you're after. Also, don't forget to put back the $\frac{1}{4}$-factor you stripped away at the beginning.
A: You should multiply also by the derivative of $(20-x^2)$, which is equal to $-2x$. Let's try this.
A: Yes,chain rule has to be applied because there are two functions here...a function of a function.
To derive that you find the derivative of the inner function multiplied by the derivative of the outer function. Using methods from Pure Mathematics 1 by Hugh Neil--->
Let $u=20-x^2$ right?
So the function is now $y=u^{0.5}$ 
So you can differentiate that which is in terms of variable u so it is $dy/du$.
since $u=20-x^2$ you differentiate that too to get $du/dx$
.But your question is asking for $dy/dx$ which can be arrived at by multiplying the two hence--->
$dy/dx = (dy/du) * (du/dx)$. If you write that out notice that the $du$ cancels out the other and you will remain with $dy/dx$.
Write it on paper and substitute,my formatting is not good need to get all my HTML tags together. Tell us if this explanation is of use to you.
