Author Topic: derivatives (Read 612 times) Tweet Share

Zahta · « **on:** May 08, 2012, 05:52:50 pm »

having trouble with these show questions on derivate, i used chain rule but i get different answers

DisaFear · « **Reply #1 on:** May 08, 2012, 06:13:21 pm »

Notice that we are differentiating in respect to x, as seen by ( $\frac{d}{dx}<br />$ ), so we can treat 'a' as a constant, just say it is '1' or '2' or something.

$y=(x^{2}\pm a^{2})$

So, in $\frac{d}{dx}(x^{2}\pm a^{2})$ , let $u=x^{2}\pm a^{2}$

From Chain Rule, we know that $\frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}$

If we take the derivative of the 'u' we find we get to $\frac{du}{dx}$
$\frac{du}{dx}=2x$ (The 'a' dies because it is a constant, remember)

Now let's find $\frac{dy}{du}$
$y=u^{\frac{1}{2}}$
$\frac{dy}{du}=\frac{1}{2}u^{\frac{-1}{2}}$

If we multiplty the two things we have calculated, we should get dy/dx (Remember to replace 'u')

$\frac{dy}{dx}=2x\times \frac{1}{2{\sqrt{x^{2}\pm a^{2}}}}$

And simplify

Keeping that in mind, try the second one. First time using Latex, tell me if I've made a mistake somewhere

Zahta · « **Reply #2 on:** May 08, 2012, 08:04:42 pm »

there is not meant to be a 2x at the top and a 2 the bottom right?? thankyou

DisaFear · « **Reply #3 on:** May 08, 2012, 08:05:57 pm »

Quote from: Zahta on May 08, 2012, 08:04:42 pm

there is not meant to be a 2x at the top and a 2 the bottom right?? thankyou

Notice the 'And simplify'. Surely you can simplify that bit XD
Does it make sense however? If you cannot do the second one, just say so, after having a go at it

Zahta · « **Reply #4 on:** May 08, 2012, 08:14:04 pm »

oh okay thankyou yeahh ill finish the rest .

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