More Differentiation Questions

**Paymemoney** · December 14th 2009, 08:05 PM

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

**pickslides** · December 14th 2009, 08:14 PM

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 $v'$

**dedust** · December 14th 2009, 08:40 PM

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 $g(x)=x^2+-a^2$

**Paymemoney** · December 14th 2009, 09:11 PM

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 $v'$

thanks got the right answer

**Paymemoney** · December 14th 2009, 09:12 PM

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.

**dedust** · December 14th 2009, 09:18 PM

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

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

**Paymemoney** · December 14th 2009, 09:37 PM

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.

**dedust** · December 14th 2009, 09:44 PM

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

**Paymemoney** · December 14th 2009, 09:46 PM

oh ok yeh i get it now

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