Hi A couple of problems i am having trouble with: 1)$\displaystyle 3\sqrt{x}(x^2+2x)$ 2)Show that $\displaystyle \frac{d}{dx}(\sqrt{x^2+-a^2})=\frac{x}{\sqrt{x^2+-a^2}}$ P.S
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Originally Posted by Paymemoney 1)$\displaystyle 3\sqrt{x}(x^2+2x)$ The product rule says For $\displaystyle y=u\times v \Rightarrow y' = u'\times v+u\times v'$ In your case make $\displaystyle u = 3\sqrt{x} $ and $\displaystyle v = x^2+2x$ Now find $\displaystyle u' $ and $\displaystyle v'$
Originally Posted by Paymemoney Hi A couple of problems i am having trouble with: 2)Show that $\displaystyle \frac{d}{dx}(\sqrt{x^2+-a^2})=\frac{x}{\sqrt{x^2+-a^2}}$ P.S $\displaystyle \frac{d}{dx}{f(g(x))}=f'(g(x)) \times g'(x)$ in this case, let $\displaystyle f(x)=\sqrt{x}$ and $\displaystyle g(x)=x^2+-a^2$
Originally Posted by pickslides The product rule says For $\displaystyle y=u\times v \Rightarrow y' = u'\times v+u\times v'$ In your case make $\displaystyle u = 3\sqrt{x} $ and $\displaystyle v = x^2+2x$ Now find $\displaystyle u' $ and $\displaystyle v'$ thanks got the right answer
Originally Posted by dedust $\displaystyle \frac{d}{dx}{f(g(x))}=f'(g(x)) \times g'(x)$ in this case, let $\displaystyle f(x)=\sqrt{x}$ and $\displaystyle g(x)=x^2+-a^2$ i can't seen to get it correct.
$\displaystyle f'(x)=\frac{1}{2\sqrt{x}}$ $\displaystyle g'(x)=2x$ then $\displaystyle f'(g(x)) \times g'(x) = . . . $
isn't $\displaystyle g'(x) = 2x +- 2a$ This is how i approached it: $\displaystyle =\frac{1}{2\sqrt(x^2)(2x)}$ $\displaystyle =\frac{1}{2\sqrt{2x^3}}$ This is where i get stuck i i don't know what can be done.
a is a constant, it doesn't depend on $\displaystyle x$, so the derivative is $\displaystyle 0$. now you should be able to calculate $\displaystyle f'(g(x)) \times g'(x)$
oh ok yeh i get it now
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