1. ## Calculus Derivative.

1) $f(x) = (x-3)x^\frac{1}{2}$ <--- to the one half power.

2) $g(x) = (x^2-3x)^-2 + 2$ <---- to the -2 power plus 2

3) $h(x) = (9-x^2)^2$

Thanks for any help.

1) $f(x) = (x^1-3)x^{\frac{1}{2}}$

$f(x) =x^{\frac{3}{2}}-3x^{\frac{1}{2}}$

Now, differentiate, got it ???

2) $g(x) = \frac{1}{(x^2+3x)^2}+2$

$g(x) = \frac{1}{x^4+6x^3+9x^2}+2$

Now, differentiate, using Quotient Rule, got it ???

3) $h(x) = (9-x^2)^2$

$
h(x) = 81-18x^2+x^4$

Now, differentiate, got it ???

3. Originally Posted by Shyam

3) $h(x) = (9-x^2)^2$

$
h(x) = 81-18x^2+x^4$

Now, differentiate, got it ???
Or use the chain rule

$\frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$

4. Originally Posted by Mathstud28
Or use the chain rule

$\frac{d}{dx}\bigg[f(g(x))\bigg]=f'(g(x))\cdot{g'(x)}$

When I use the chain rule for 3.

Would it be $2(9-x^2)(-2x)$

= $-4x(9-x^2)$

5. Originally Posted by teddybear
When I use the chain rule for 3.

Would it be $2(9-x^2)(-2x)$

= $-4x(9-x^2)$
that's correct

6. I think I finished it... but I'm not sure if I did it right.. Can someone check for me?

1. ${\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}$

For number 2, I have a problem. After I used the quotient rule, I came upon
$-4x^3-18x^2-18x/(x^4+6x^3+9x^2)^2$ and I don't know how to simplify that.

7. Originally Posted by teddybear
I think I finished it... but I'm not sure if I did it right.. Can someone check for me?

1. ${\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}$
yes
For number 2, I have a problem. After I used the quotient rule, I came upon
$-4x^3-18x^2-18x/(x^4+6x^3+9x^2)^2$ and I don't know how to simplify that.
why use the quotient rule? as the problem was written, it seems to suggest you should use the chain rule. much in the same way you did problem 3

8. Originally Posted by teddybear
1) $f(x) = (x-3)x^\frac{1}{2}$ <--- to the one half power.

2) $g(x) = (x^2-3x)^-2 + 2$ <---- to the -2 power plus 2

3) $h(x) = (9-x^2)^2$

Thanks for any help.
Yes, the chain rule is easier for 2).

$g = (x^2 - 3x)^{-2} + 2.$

Let $u = x^2 - 3x$ so $g = u^{-2} + 2.$

$\frac{du}{dx} = 2x, \frac{dg}{du} = -2u^{-3} = -2(x^2 - 3x)^{-3}$

So $\frac{dg}{dx} = \frac{du}{dx}\times\frac{dg}{du} = -4x(x^2 - 3x)^{-3}$.