1. ## More Differentiation Questions

Hi
A couple of problems i am having trouble with:

1) $3\sqrt{x}(x^2+2x)$
2)Show that $\frac{d}{dx}(\sqrt{x^2+-a^2})=\frac{x}{\sqrt{x^2+-a^2}}$

P.S

2. Originally Posted by Paymemoney
1) $3\sqrt{x}(x^2+2x)$

The product rule says

For $y=u\times v \Rightarrow y' = u'\times v+u\times v'$

In your case make $u = 3\sqrt{x}$ and $v = x^2+2x$

Now find $u'$ and $v'$

3. Originally Posted by Paymemoney
Hi
A couple of problems i am having trouble with:

2)Show that $\frac{d}{dx}(\sqrt{x^2+-a^2})=\frac{x}{\sqrt{x^2+-a^2}}$

P.S
$\frac{d}{dx}{f(g(x))}=f'(g(x)) \times g'(x)$

in this case, let $f(x)=\sqrt{x}$ and $g(x)=x^2+-a^2$

4. Originally Posted by pickslides
The product rule says

For $y=u\times v \Rightarrow y' = u'\times v+u\times v'$

In your case make $u = 3\sqrt{x}$ and $v = x^2+2x$

Now find $u'$ and $v'$

5. Originally Posted by dedust
$\frac{d}{dx}{f(g(x))}=f'(g(x)) \times g'(x)$

in this case, let $f(x)=\sqrt{x}$ and $g(x)=x^2+-a^2$
i can't seen to get it correct.

6. $f'(x)=\frac{1}{2\sqrt{x}}$
$g'(x)=2x$

then
$f'(g(x)) \times g'(x) = . . .$

7. isn't $g'(x) = 2x +- 2a$

This is how i approached it:
$=\frac{1}{2\sqrt(x^2)(2x)}$
$=\frac{1}{2\sqrt{2x^3}}$

This is where i get stuck i i don't know what can be done.

8. a is a constant, it doesn't depend on $x$, so the derivative is $0$.
now you should be able to calculate $f'(g(x)) \times g'(x)$

9. oh ok yeh i get it now