1. ## Derivative

going through practice problems for a test

Am I doing this correctly?

Find the differential:

$y=\sqrt{9-x^2}=(9-x^2)^\frac{1}{2}$

via power rule: $y'=\frac{1}{2}(9-x^2)^\frac{-1}{2}$

and so,

$dy=\frac{1}{\sqrt{4.5-.5x^2}}*dx$

2. Originally Posted by makieru
going through practice problems for a test

Am I doing this correctly?

Find the differential:

$y=\sqrt{9-x^2}=(9-x^2)^\frac{1}{2}$

via power rule: $y'=\frac{1}{2}(9-x^2)^\frac{-1}{2}$

and so,

$dy=\frac{1}{\sqrt{4.5-.5x^2}}*dx$

You need the power rule AND the chain rule...

Let $\sqrt{x}$= f(x) and $9-x^2=g(x)$

Then, $y=f(g(x))\Rightarrow \frac{dy}{dx}=f'(g(x))\cdot g'(x)$ (this is just the chain rule)

So in your case, you take the derivative of f correctly, but never multiply by the derivative of g

Also, that $\frac{1}{2}$ isn't inside the square root, so you can't multiply $9-x^2$ by $\frac{1}{2}$ under the radical

3. Originally Posted by makieru
going through practice problems for a test

Am I doing this correctly?

Find the differential:

$y=\sqrt{9-x^2}=(9-x^2)^\frac{1}{2}$

via power rule: $y'=\frac{1}{2}(9-x^2)^\frac{-1}{2}$

$\color{red}\mbox{This is wrong: you forgot to multiply by the inner derivative}-2x$

and so,

$dy=\frac{1}{\sqrt{4.5-.5x^2}}*dx$